Chapter 9 — Applications of Derivatives

9.1 What Derivatives Tell Us

For three chapters you have learned to compute the derivative — the limit rules, the chain rule, implicit differentiation. Now comes the payoff. A derivative is not merely a number attached to a point; it is a report on how a function behaves there. Read that report carefully and you can reconstruct the entire shape of a curve without plotting a single point by brute force.

Here is the menu of questions the derivative answers:

• Where is $f$ increasing or decreasing?
• Where are the peaks and valleys — the maxima and minima?
• Does the curve open upward like a bowl or downward like a dome — its concavity?
• Where does the curvature switch direction — its inflection points?
• How do stubborn limits of the form $0/0$ or $\infty/\infty$ actually behave?

Every one of these is answered by $f'$ and $f''$. This chapter assembles the toolkit, and beneath all of it sits one theorem — the Mean Value Theorem — that we will not merely state but prove, because nearly every fact in this chapter is secretly a corollary of it.

The Key Insight. A function's shape is encoded in its derivatives. The first derivative $f'$ governs slope — and therefore monotonicity and the location of extrema. The second derivative $f''$ governs curvature — and therefore concavity and inflection points. Master the dialogue between $f'$ and $f''$ and you can read any curve like a sentence.

This is the second of our recurring themes made vivid: geometry and algebra are inseparable. Each sign of $f'$ is a direction the curve travels; each sign of $f''$ is a way the curve bends. We will never write a formula without drawing its picture.

Warning. Optimization word problems — "what dimensions minimize the cost of a can?" — get a dedicated treatment in Chapter 10. This chapter builds the machinery (critical points, the derivative tests); Chapter 10 turns it loose on the real world. When you see optimization previewed here, it is a trailer, not the feature.

9.2 Absolute Extrema and the Extreme Value Theorem

Before classifying local bumps, we settle the global question: does a function even have a highest and lowest value?

A function $f$ has an absolute maximum (or global maximum) at $c$ if $f(c) \ge f(x)$ for every $x$ in the domain. Replace $\ge$ with $\le$ for an absolute minimum. These are the literal top and bottom of the entire graph. By contrast, a local maximum at $c$ only requires $f(c) \ge f(x)$ for $x$ near $c$ — a hilltop, not necessarily the highest mountain.

Absolute extrema need not exist. The function $f(x) = x$ on all of $\mathbb{R}$ has neither; it runs off to $\pm\infty$. Even on a bounded but open interval things can fail: $f(x) = x$ on $(0,1)$ gets arbitrarily close to $1$ but never attains it. The cure is a closed interval plus continuity.

Extreme Value Theorem. If $f$ is continuous on a closed, bounded interval $[a,b]$, then $f$ attains an absolute maximum and an absolute minimum somewhere on $[a,b]$.

Both hypotheses are load-bearing. Drop continuity and a jump can let the function approach a value it never reaches. Drop closedness and the endpoint can be the missing supremum. The proof is a genuine theorem of real analysis (it rests on the completeness of $\mathbb{R}$), and we flag it as optional.

Math Major Sidebar. The Extreme Value Theorem is not provable by calculus alone — it is a topological fact: a continuous image of a compact set is compact, and a compact subset of $\mathbb{R}$ is closed and bounded, hence contains its supremum and infimum. The first careful proofs came from Weierstrass and Bolzano in the 19th century, long after Newton and Leibniz used the result freely. If you continue to real analysis, you will prove it from the Bolzano–Weierstrass theorem. For now, accept it: continuity on a closed interval guarantees a true max and min.

9.3 Critical Points and Fermat's Theorem

So where do extrema hide? The answer is beautifully restrictive.

A critical point of $f$ is a number $c$ in the domain where either $f'(c) = 0$ or $f'(c)$ does not exist. These are the only places — apart from the endpoints of a closed interval — where extrema can occur.

Fermat's Theorem. If $f$ has a local extremum at an interior point $c$ and $f'(c)$ exists, then $f'(c) = 0$.

The intuition is irresistible. At the top of a smooth hill, the tangent line is horizontal — if it tilted upward, you could climb higher to the right; if downward, higher to the left. A genuine local maximum leaves nowhere higher to go, so the slope must be zero. The same picture, inverted, holds at a valley.

A quick proof. Suppose $f$ has a local maximum at $c$. For small $h > 0$, $f(c+h) - f(c) \le 0$, so the right-hand difference quotient $\frac{f(c+h)-f(c)}{h} \le 0$, and its limit gives $f'(c) \le 0$. For small $h < 0$, the numerator is still $\le 0$ but now $h < 0$, so the quotient is $\ge 0$, giving $f'(c) \ge 0$. The only number that is both $\le 0$ and $\ge 0$ is $0$. Hence $f'(c) = 0$.

Two warnings keep this honest.

Common Pitfall. Many students believe "critical point" means "$f'(c) = 0$." But the derivative-doesn't-exist case matters just as much. The function $f(x) = |x|$ has a local (and absolute) minimum at $x = 0$, yet $f'(0)$ does not exist — there is a corner, not a horizontal tangent. If you only solve $f'(x) = 0$, you will miss extrema at corners, cusps, and vertical tangents.

Warning. Fermat's theorem is a one-way street: a critical point is a candidate for an extremum, not a guarantee. The classic counterexample is $f(x) = x^3$, with $f'(0) = 0$. The slope vanishes at the origin, yet the curve sails straight through — increasing on both sides, no extremum at all. Critical points are suspects; you still need a test to convict them.

So the strategy for finding absolute extrema on a closed interval $[a,b]$ is now complete — the Closed Interval Method:

1. Find all critical points of $f$ in the open interval $(a,b)$.
2. Evaluate $f$ at each critical point and at both endpoints $a$ and $b$.
3. The largest of these values is the absolute maximum; the smallest is the absolute minimum.

The Extreme Value Theorem promises the answer exists; Fermat's theorem promises this short list contains it.

Worked example. Find the absolute extrema of $f(x) = x^3 - 3x$ on $[-2, 2]$.

The derivative is $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$, giving critical points $x = -1$ and $x = 1$ (both interior). Evaluate the candidates:

$$f(-2) = -8 + 6 = -2,\quad f(-1) = -1+3 = 2,\quad f(1) = 1 - 3 = -2,\quad f(2) = 8 - 6 = 2.$$

The absolute maximum is $2$ (attained at both $x = -1$ and $x = 2$); the absolute minimum is $-2$ (attained at both $x = -2$ and $x = 1$). Notice that the extreme values can be tied, and that endpoints competed on equal footing with critical points.

Check Your Understanding. Find the absolute maximum and minimum of $f(x) = x^4 - 8x^2$ on $[-1, 3]$.

Answer

$f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2)$, so the interior critical points in $[-1,3]$ are $x = 0$ and $x = 2$ (the point $x = -2$ lies outside). Evaluate: $f(-1) = 1 - 8 = -7$, $f(0) = 0$, $f(2) = 16 - 32 = -16$, $f(3) = 81 - 72 = 9$. Absolute max $= 9$ at $x = 3$; absolute min $= -16$ at $x = 2$.

9.4 Monotonicity: The Increasing/Decreasing Test

A function is increasing on an interval if $f(b) > f(a)$ whenever $b > a$ in that interval, and decreasing if $f(b) < f(a)$ whenever $b > a$. A function that is increasing or decreasing throughout an interval is called monotonic there.

The derivative diagnoses monotonicity directly.

Increasing/Decreasing Test. Suppose $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. - If $f'(x) > 0$ for all $x \in (a,b)$, then $f$ is increasing on $[a,b]$. - If $f'(x) < 0$ for all $x \in (a,b)$, then $f$ is decreasing on $[a,b]$. - If $f'(x) = 0$ for all $x \in (a,b)$, then $f$ is constant on $[a,b]$.

This feels obvious — positive slope means the curve heads uphill — but the honest proof requires the Mean Value Theorem, which is exactly why we will prove the MVT in §9.7. Hold that thought; the test is so useful we use it now and justify it shortly.

Worked example. Where is $f(x) = x^3 - 3x^2 + 4$ increasing and decreasing?

Differentiate: $f'(x) = 3x^2 - 6x = 3x(x-2)$. The roots are $x = 0$ and $x = 2$, dividing the line into three intervals. Test a point in each:

The curve rises to $x = 0$, dips down to $x = 2$, then rises again — exactly the silhouette of a cubic with two turning points. This sign chart is the single most useful diagram in differential calculus; practice drawing it.

Geometric Intuition. Picture a marble rolling along the graph from left to right. Where $f' > 0$ the marble climbs; where $f' < 0$ it descends; where $f' = 0$ it pauses at a crest or trough. A local maximum is the instant the marble stops climbing and begins to fall — precisely where $f'$ switches from $+$ to $-$. That switch is the whole content of the first derivative test, which is next.

9.5 The First Derivative Test

Combine Fermat's theorem (extrema live at critical points) with the Increasing/Decreasing Test (sign of $f'$ tells direction) and you get a complete classification.

First Derivative Test. Let $c$ be a critical point of a continuous function $f$. Examine the sign of $f'$ just to the left and just to the right of $c$: - $f'$ changes from $+$ to $-$ at $c$ $\;\Longrightarrow\;$ local maximum at $c$ (the curve stops rising and starts falling). - $f'$ changes from $-$ to $+$ at $c$ $\;\Longrightarrow\;$ local minimum at $c$ (the curve stops falling and starts rising). - $f'$ does not change sign at $c$ $\;\Longrightarrow\;$ neither — a horizontal-tangent flex like $x^3$ at the origin.

The beauty of this test is that it never asks for the second derivative and never fails to give an answer — it always works, even at corners where $f''$ does not exist.

Worked example. Classify the critical points of $f(x) = x^3 - 3x$.

We found $f'(x) = 3(x-1)(x+1)$ with critical points $x = \pm 1$. The sign chart:

At $x = -1$, $f'$ goes $+ \to -$: a local maximum, with $f(-1) = 2$. At $x = 1$, $f'$ goes $- \to +$: a local minimum, with $f(1) = -2$. The cubic has a hill at $(-1, 2)$ and a valley at $(1, -2)$, consistent with our absolute-extrema work in §9.3.

9.6 Concavity, the Second Derivative Test, and Inflection Points

The first derivative tells you which way the curve is heading; the second derivative tells you how it bends.

A function is concave up on an interval if its graph lies above each of its tangent lines there — it cups upward like a bowl. It is concave down if the graph lies below its tangents — it caps downward like a dome. The link to $f''$ is direct: concave up means the slope $f'$ is increasing, which means $(f')' = f'' > 0$.

Concavity Test. - If $f''(x) > 0$ on an interval, then $f$ is concave up there. - If $f''(x) < 0$ on an interval, then $f$ is concave down there.

Geometric Intuition. Concave-up is a smile, concave-down is a frown — the classic mnemonic, and it is exactly right. But here is the deeper picture: concavity is the rate of change of slope. On a concave-up stretch the tangent lines keep tilting more steeply upward (or less steeply downward) as you move right; the curve is always pulling away above its current tangent. That is why a tangent line underestimates a concave-up function — a fact we will exploit in Chapter 11 when we study linear approximation, and again in Chapter 10 for optimization.

This curvature information sharpens extremum classification into a one-shot test.

Second Derivative Test. Suppose $f'(c) = 0$ (so $c$ is a critical point with a horizontal tangent) and $f''$ is continuous near $c$: - $f''(c) > 0$ $\;\Longrightarrow\;$ local minimum at $c$ (sitting at the bottom of a concave-up bowl). - $f''(c) < 0$ $\;\Longrightarrow\;$ local maximum at $c$ (sitting at the top of a concave-down dome). - $f''(c) = 0$ $\;\Longrightarrow\;$ inconclusive — the test says nothing; fall back on the first derivative test.

Worked example. Reclassify the critical points of $f(x) = x^3 - 3x$ using $f''$.

Here $f''(x) = 6x$. At $x = -1$: $f''(-1) = -6 < 0$, a local maximum. At $x = 1$: $f''(1) = 6 > 0$, a local minimum. Same verdict as the first derivative test, reached faster — one evaluation instead of a sign chart.

Common Pitfall. When $f''(c) = 0$, students often conclude "inflection point" or "no extremum." Both can be wrong. Consider $f(x) = x^4$: $f'(0) = 0$ and $f''(0) = 0$, yet the origin is plainly an absolute minimum (the function is positive everywhere else). The second derivative test is merely inconclusive at $x = 0$; the first derivative test ($f'$ goes $- \to +$) correctly finds the minimum. A zero second derivative is a question, not an answer.

An inflection point is a point on the graph where the concavity actually changes — where $f''$ switches sign. Two cautions: $f''(c) = 0$ is necessary but not sufficient (again $x^4$ at $0$: $f''=0$ but no sign change, hence no inflection); and an inflection can also occur where $f''$ fails to exist, provided the concavity genuinely flips across that point.

Worked example. For $f(x) = x^3$, $f''(x) = 6x$ is negative for $x < 0$ (concave down) and positive for $x > 0$ (concave up). The concavity flips at $x = 0$, so $(0,0)$ is an inflection point — even though it is not an extremum. For $f(x) = \sin x$, $f''(x) = -\sin x$ changes sign at every $x = n\pi$, so the sine curve has inflection points at $(n\pi, 0)$ for each integer $n$, exactly where the wave crosses its midline.

Check Your Understanding. Find the intervals of concavity and the inflection points of $f(x) = x^4 - 6x^2$.

Answer

$f''(x) = 12x^2 - 12 = 12(x^2 - 1) = 12(x-1)(x+1)$. This is positive for $|x| > 1$ (concave up on $(-\infty,-1)$ and $(1,\infty)$) and negative for $|x| < 1$ (concave down on $(-1,1)$). Concavity changes at both $x = -1$ and $x = 1$, so the inflection points are $(-1, -5)$ and $(1, -5)$ (since $f(\pm 1) = 1 - 6 = -5$).

9.7 The Mean Value Theorem

We now reach the theoretical heart of the chapter. Everything above — the Increasing/Decreasing Test, the very meaning of "$f' > 0$ implies increasing" — rests on one theorem that we have been quietly borrowing. It is time to pay the debt.

Mean Value Theorem (MVT). If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one point $c \in (a,b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$

In words: somewhere strictly inside the interval, the instantaneous rate of change equals the average rate of change over the whole interval. The right-hand side is the slope of the secant line joining $\big(a, f(a)\big)$ to $\big(b, f(b)\big)$; the left-hand side is the slope of the tangent at $c$. So the MVT says: somewhere, the tangent runs exactly parallel to the secant.

Real-World Application — Speeding tickets (kinematics). Drive 120 miles in 2 hours and your average speed is 60 mph. The MVT guarantees that at some instant your speedometer read exactly 60 mph — your instantaneous speed must equal your average speed at least once. This is the mathematical principle behind average-speed enforcement cameras on highways: two timestamped photos at a known distance apart prove, by the MVT, that you exceeded the limit at some moment, even though no single radar gun ever clocked you. The theorem turns an average into a certainty about an instant.

9.7.1 Rolle's Theorem: the special case

Following the pilot's "Key Idea first" structure, we prove the MVT in two stages. The engine is a deceptively simple special case.

Rolle's Theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists $c \in (a,b)$ with $f'(c) = 0$.

Why we care. Rolle's theorem is the MVT in the case where the secant line is horizontal. If a smooth curve starts and ends at the same height, it must turn around somewhere in between — and at that turning point the tangent is flat. Get Rolle's, and the full MVT is a one-line tilt away.

Key Idea. A continuous function on a closed interval attains a max and a min (Extreme Value Theorem). If both occur at the endpoints, the function is constant and $f' = 0$ everywhere. Otherwise an extremum lands at an interior point — and by Fermat's theorem the derivative there is zero.

Formal proof. Since $f$ is continuous on the closed interval $[a,b]$, the Extreme Value Theorem (§9.2) guarantees $f$ attains an absolute maximum value $M$ and an absolute minimum value $m$ on $[a,b]$.

Case 1: $M = m$. Then $f$ is constant on $[a,b]$, so $f'(x) = 0$ for every interior $x$. Any $c \in (a,b)$ works.

Case 2: $M \ne m$. At least one of these extreme values differs from the common endpoint value $f(a) = f(b)$. Say it is the maximum $M$ (the argument for the minimum is identical). Then $M$ is attained at some point $c$ that is not an endpoint, so $c \in (a,b)$. Because $f$ is differentiable on $(a,b)$, $f'(c)$ exists, and since $f$ has a local maximum at the interior point $c$, Fermat's theorem (§9.3) gives $f'(c) = 0$. $\blacksquare$

What this means. Rolle's theorem is the precise statement of "what goes up and comes back must level off." It is pure existence: it tells you a flat tangent exists without telling you where. Its hypotheses are non-negotiable — remove differentiability and $f(x) = |x|$ on $[-1,1]$ satisfies $f(-1)=f(1)$ yet never has a horizontal tangent (the corner at $0$ is the culprit).

9.7.2 Proving the Mean Value Theorem

Why we care. The MVT is the bridge from instantaneous to average. It is what lets us conclude global behavior ("$f$ is increasing on the whole interval") from local information ("$f' > 0$ at each point"). Without it, the Increasing/Decreasing Test, the uniqueness of antiderivatives up to a constant, and the Fundamental Theorem of Calculus (Chapter 14) would all collapse.

Key Idea. Tilt the picture. The MVT looks like Rolle's theorem with the interval lifted onto a slant. So subtract off the secant line — manufacture a new function whose endpoints are level — and Rolle's theorem applies to it.

Formal proof. Let $L(x)$ be the secant line through $\big(a, f(a)\big)$ and $\big(b, f(b)\big)$:

$$L(x) = f(a) + \frac{f(b) - f(a)}{b - a}\,(x - a).$$

Define the auxiliary function $g(x) = f(x) - L(x)$, the gap between the curve and its secant. Then $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$, because both $f$ and the line $L$ are. At the endpoints the curve meets the secant, so

$$g(a) = f(a) - L(a) = 0 \qquad\text{and}\qquad g(b) = f(b) - L(b) = 0.$$

Thus $g(a) = g(b)$, and $g$ satisfies every hypothesis of Rolle's theorem. Therefore there exists $c \in (a,b)$ with $g'(c) = 0$. But

$$g'(x) = f'(x) - L'(x) = f'(x) - \frac{f(b) - f(a)}{b - a},$$

since the slope of the line $L$ is the constant secant slope. Setting $g'(c) = 0$ gives

$$f'(c) = \frac{f(b) - f(a)}{b - a}. \qquad \blacksquare$$

What this means. The MVT is Rolle's theorem viewed on a slope. Geometrically, you slide the secant line upward (keeping its slope) until it just kisses the curve tangentially; the point of tangency is $c$. The proof's one clever move — subtracting the secant — is the same trick you will see again and again in analysis: reduce a tilted problem to a flat one you have already solved.

9.7.3 Consequences of the MVT

Now we collect the debts. Each of these "obvious" facts is a direct corollary of the MVT.

Increasing/Decreasing Test (proof). Suppose $f' > 0$ on $(a,b)$. Take any two points $x_1 < x_2$ in the interval. By the MVT applied to $[x_1, x_2]$, there is a $c$ between them with $$f(x_2) - f(x_1) = f'(c)\,(x_2 - x_1).$$ The factor $f'(c) > 0$ and $x_2 - x_1 > 0$, so $f(x_2) - f(x_1) > 0$, i.e. $f(x_2) > f(x_1)$. Hence $f$ is increasing. The decreasing case is identical with the inequality flipped. This is the proof we owed in §9.4.

Constant Function Theorem. If $f'(x) = 0$ for all $x$ in an interval, then $f$ is constant there. (Same argument: $f(x_2) - f(x_1) = f'(c)(x_2 - x_1) = 0$ for any two points.)

Functions with equal derivatives differ by a constant. If $f'(x) = g'(x)$ on an interval, then $f(x) = g(x) + C$ for some constant $C$. (Apply the Constant Function Theorem to $f - g$.) This is the fact that justifies the "$+C$" in every indefinite integral and underlies the Fundamental Theorem of Calculus in Chapter 14.

Math Major Sidebar. Notice the dependency chain: Extreme Value Theorem $\Rightarrow$ Fermat $\Rightarrow$ Rolle $\Rightarrow$ MVT $\Rightarrow$ (Increasing/Decreasing Test, Constant Function Theorem, uniqueness of antiderivatives) $\Rightarrow$ FTC. The Mean Value Theorem is the structural keystone of single-variable calculus. It also generalizes: Cauchy's Mean Value Theorem ($\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}$) is the engine behind L'Hôpital's rule in §9.8, and Taylor's theorem with the Lagrange remainder (Chapter 23) is the MVT carried out to higher order. When you meet those later, remember they are all the same horizontal-tangent idea, tilted and iterated.

Worked example. Verify the MVT for $f(x) = x^2$ on $[0, 2]$. The average rate of change is $\frac{f(2)-f(0)}{2-0} = \frac{4-0}{2} = 2$. We need $c$ with $f'(c) = 2c = 2$, so $c = 1 \in (0,2)$. The tangent at $x=1$ has slope $2$, parallel to the secant from $(0,0)$ to $(2,4)$. ✓

9.8 L'Hôpital's Rule

We turn from theory to a powerful computational tool for limits that resist every earlier trick.

An indeterminate form is a limit whose naive substitution yields an ambiguous expression — $\frac{0}{0}$ or $\frac{\infty}{\infty}$ — where the answer genuinely depends on how the numerator and denominator race to their limits. L'Hôpital's rule converts the race of values into a race of slopes.

L'Hôpital's Rule. Suppose $f$ and $g$ are differentiable near $a$ (except possibly at $a$), with $g'(x) \ne 0$ near $a$. If $$\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0 \qquad\text{or}\qquad \lim_{x \to a} f(x) = \pm\infty,\ \lim_{x \to a} g(x) = \pm\infty,$$ then $$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},$$ provided the limit on the right exists (or is $\pm\infty$). The rule also holds for one-sided limits and for $a = \pm\infty$.

Geometric Intuition. Why should differentiating top and bottom help? Near a point where both $f$ and $g$ vanish, each is approximately linear: $f(x) \approx f'(a)(x-a)$ and $g(x) \approx g'(a)(x-a)$ by the tangent-line approximation. The common factor $(x-a)$ cancels in the ratio, leaving $f'(a)/g'(a)$. L'Hôpital's rule is the rigorous version of "replace each function by its tangent line and cancel." The genuine proof uses Cauchy's Mean Value Theorem from the sidebar above.

9.8.1 The basic forms

Example 1. $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$. Both numerator and denominator tend to $0$ — form $\frac{0}{0}$. Differentiate each: $\lim_{x\to 0} \frac{\cos x}{1} = \cos 0 = 1$. This recovers the famous limit we proved geometrically in Chapter 3, now in one line. (A subtle point: this is not circular only because we already know $\frac{d}{dx}\sin x = \cos x$ independently.)

Example 2. $\displaystyle\lim_{x \to \infty} \frac{e^x}{x^{100}}$. Form $\frac{\infty}{\infty}$. Each application of L'Hôpital lowers the polynomial's degree by one while leaving $e^x$ untouched. After $100$ applications, $\lim_{x\to\infty} \frac{e^x}{100!} = \infty$. The lesson is permanent: exponential growth outpaces every polynomial, no matter how large the degree.

Example 3. $\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^3}$. Form $\frac{0}{0}$. Apply repeatedly, checking that each stage is still indeterminate: $$\lim_{x\to 0}\frac{x - \sin x}{x^3} = \lim_{x\to 0}\frac{1 - \cos x}{3x^2} = \lim_{x\to 0}\frac{\sin x}{6x} = \lim_{x\to 0}\frac{\cos x}{6} = \frac{1}{6}.$$ Three applications, each legitimate because the previous result was $\frac{0}{0}$. The fourth substitution gives $\cos 0 / 6 = 1/6$.

Check Your Understanding. Evaluate $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$.

Answer

Substitution gives $\frac{0}{0}$. Apply L'Hôpital: $\lim_{x\to 0}\frac{e^x - 1}{2x}$, still $\frac{0}{0}$. Apply again: $\lim_{x\to 0}\frac{e^x}{2} = \frac{1}{2}$. (You can preview Chapter 23: the Taylor series $e^x = 1 + x + \tfrac{x^2}{2} + \cdots$ makes the numerator $\tfrac{x^2}{2}+\cdots$, so the ratio tends to $\tfrac12$ — same answer, deeper reason.)

9.8.2 The other indeterminate forms

L'Hôpital's rule only applies to $\frac{0}{0}$ and $\frac{\infty}{\infty}$. Every other indeterminate form must first be algebraically reshaped into one of those two.

• $0 \cdot \infty$: rewrite the product as a quotient, $f \cdot g = \dfrac{f}{1/g}$, to get $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
• $\infty - \infty$: combine over a common denominator (or rationalize) into a single fraction.
• $0^0$, $1^\infty$, $\infty^0$: these are exponential indeterminates. Take the natural logarithm, evaluate the resulting limit, then exponentiate.

Example 4 (form $0 \cdot \infty$). $\displaystyle\lim_{x \to 0^+} x \ln x$. As $x \to 0^+$, $x \to 0$ and $\ln x \to -\infty$ — form $0 \cdot (-\infty)$. Rewrite as a quotient: $$\lim_{x\to 0^+} \frac{\ln x}{1/x} \quad\text{(form } \tfrac{-\infty}{\infty}\text{)} \;=\; \lim_{x\to 0^+}\frac{1/x}{-1/x^2} = \lim_{x\to 0^+}(-x) = 0.$$

Example 5 (form $0^0$). $\displaystyle\lim_{x \to 0^+} x^x$. Let $y = x^x$, so $\ln y = x \ln x$. From Example 4, $\lim_{x\to 0^+}\ln y = 0$, hence $\lim_{x\to 0^+} y = e^0 = 1$. So $x^x \to 1$ — a clean and slightly surprising result.

Real-World Application — Continuous compounding (economics/finance). The constant $e$ itself is an indeterminate-form limit: $\lim_{n\to\infty}\left(1 + \tfrac{r}{n}\right)^n$, a form $1^\infty$. Taking logs turns it into $\lim_{n\to\infty} n\ln(1+\tfrac{r}{n})$, a $0\cdot\infty$ form that L'Hôpital evaluates to $r$, giving $e^r$ — the formula for continuously compounded interest. Every bank's continuous-interest model and every exponential-growth law in epidemiology rests on resolving this single indeterminate form.

9.8.3 When L'Hôpital fails or misleads

Common Pitfall. The number-one L'Hôpital error is applying it to a limit that is not indeterminate. Consider $\lim_{x\to 1}\frac{x+1}{x^2+1} = \frac{2}{2} = 1$ by direct substitution — there is nothing indeterminate here. Differentiating top and bottom would give $\lim_{x\to 1}\frac{1}{2x} = \frac{1}{2}$, the wrong answer. Always verify the form is $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before differentiating.

Warning. L'Hôpital can fail even on a legitimate $\frac{\infty}{\infty}$ form when the ratio of derivatives has no limit. Take $\lim_{x\to\infty}\frac{x + \sin x}{x}$. The form is $\frac{\infty}{\infty}$, and the true value is $1$ (since $\frac{\sin x}{x}\to 0$). But L'Hôpital gives $\lim_{x\to\infty}\frac{1 + \cos x}{1}$, which oscillates forever and does not exist. The rule's hypothesis — "provided the limit on the right exists" — was violated, so the rule simply does not apply. When the differentiated limit misbehaves, abandon L'Hôpital and return to algebra; the original limit may exist perfectly well.

9.9 Curve Sketching: Putting It All Together

We can now read a curve's complete shape from calculus alone. The systematic procedure:

1. Domain. Where is $f$ defined? Exclude division-by-zero and even roots of negatives.
2. Intercepts. $y$-intercept is $f(0)$; $x$-intercepts solve $f(x) = 0$.
3. Symmetry. Even ($f(-x)=f(x)$, mirror across the $y$-axis), odd ($f(-x)=-f(x)$, symmetric through the origin), or periodic?
4. Asymptotes. Vertical (where $f \to \pm\infty$), horizontal ($\lim_{x\to\pm\infty} f$), oblique (when a rational function's numerator degree is one more than the denominator's — see §9.10).
5. First derivative. Find critical points; build the sign chart of $f'$ for intervals of increase/decrease.
6. Local extrema. Classify each critical point (first or second derivative test).
7. Second derivative. Sign chart of $f''$ for concavity; locate inflection points.
8. Assemble. Plot the special points and connect them honoring slope and concavity.

Worked example — a cubic. Sketch $f(x) = x^3 - 3x^2 + 2$.

Domain: all of $\mathbb{R}$. Intercepts: $f(0) = 2$. For the $x$-intercepts, notice $f(1) = 1 - 3 + 2 = 0$, so $x = 1$ is a root; factoring out $(x-1)$ gives $x^3 - 3x^2 + 2 = (x-1)(x^2 - 2x - 2)$, and the quadratic formula gives the other roots $x = 1 \pm \sqrt{3} \approx -0.73$ and $2.73$. Symmetry: none (mixing even and odd powers). Asymptotes: none — polynomials grow without bound but have no asymptotes.

First derivative: $f'(x) = 3x^2 - 6x = 3x(x-2)$, critical points $x = 0$ and $x = 2$. Sign chart: $f' > 0$ on $(-\infty,0)$, $f' < 0$ on $(0,2)$, $f' > 0$ on $(2,\infty)$. So a local maximum at $(0, 2)$ and a local minimum at $(2, -2)$.

Second derivative: $f''(x) = 6x - 6 = 6(x-1)$. Negative for $x < 1$ (concave down), positive for $x > 1$ (concave up), so an inflection point at $(1, 0)$ — which is also, neatly, an $x$-intercept.

Assemble: the curve climbs to the peak $(0,2)$, descends through the inflection at $(1,0)$ while concave down, bottoms out at the valley $(2,-2)$ now concave up, then climbs again. A textbook cubic silhouette.

Common Pitfall. A widespread error in this exact problem is claiming the $x$-intercepts are "approximately $-0.73,\,0.84,\,2.89$." They are not. Because $f(1) = 0$ exactly, one root is precisely $x = 1$, and the symmetric pair is $1 \pm \sqrt{3}$. Always test small integers for exact roots before resorting to a numerical solver — a single exact root often factors the whole problem.

Check Your Understanding. For $f(x) = x^3 - 3x^2 + 2$, is the inflection point a critical point?

Answer

No. The inflection point is at $x = 1$, where $f''(1) = 0$ but $f'(1) = 3(1)(1-2) = -3 \ne 0$. The curve is decreasing (nonzero slope) as it passes through the inflection; concavity changes there, but it is not a peak or valley. Inflection points and critical points are different ideas and only rarely coincide.

9.10 Asymptotes and Two Harder Sketches

For a rational function $f(x) = p(x)/q(x)$ in lowest terms:

• Vertical asymptotes occur at the zeros of the denominator $q(x)$ that are not cancelled by zeros of $p(x)$.
• Horizontal asymptotes come from the end behavior. If $\deg p < \deg q$, then $y = 0$. If $\deg p = \deg q$, then $y$ equals the ratio of leading coefficients. If $\deg p > \deg q$, there is no horizontal asymptote.
• Oblique (slant) asymptotes appear when $\deg p = \deg q + 1$. Polynomial long division writes $f(x) = (mx + b) + \frac{\text{remainder}}{q(x)}$; the line $y = mx + b$ is the slant asymptote.

Worked example — an oblique asymptote. Analyze $f(x) = \dfrac{x^2 + 1}{x - 1}$.

Long division: $x^2 + 1 = (x-1)(x+1) + 2$, so $f(x) = x + 1 + \dfrac{2}{x-1}$. As $x \to \pm\infty$ the last term vanishes, leaving the oblique asymptote $y = x + 1$. The denominator vanishes at $x = 1$ (and the numerator does not), so there is a vertical asymptote at $x = 1$.

Worked example — a bounded rational function. Sketch $f(x) = \dfrac{x}{x^2 + 1}$.

Domain: all of $\mathbb{R}$ (the denominator is never zero). Intercept: $f(0) = 0$. Symmetry: odd, since $f(-x) = -f(x)$ — symmetric through the origin. Asymptotes: as $x \to \pm\infty$, $f \to 0$, so $y = 0$ is a horizontal asymptote (no vertical asymptotes).

First derivative (quotient rule): $f'(x) = \dfrac{(x^2+1) - x(2x)}{(x^2+1)^2} = \dfrac{1 - x^2}{(x^2+1)^2}$. The sign is governed by $1 - x^2$: positive for $|x| < 1$ (increasing), negative for $|x| > 1$ (decreasing). So a local maximum at $\left(1, \tfrac12\right)$ and a local minimum at $\left(-1, -\tfrac12\right)$.

Second derivative: a computation (verified by sympy in §9.11) gives inflection points at $x = 0$ and $x = \pm\sqrt{3}$. The graph is a gentle S-curve through the origin, rising to a hump of height $\tfrac12$ at $x = 1$, then sinking back toward the axis — a shape that appears throughout signal processing as a smoothed step.

9.11 Computation: Symbolic Curve Analysis

Following our standard three-tier pattern — pose analytically, solve by hand, verify by machine — let sympy confirm the cubic from §9.9 and the rational function from §9.10.

import sympy as sp

x = sp.Symbol('x')
f = x**3 - 3*x**2 + 2

f1 = sp.diff(f, x)                 # first derivative
critical = sp.solve(f1, x)
print("Critical points:", critical)            # [0, 2]

f2 = sp.diff(f, x, 2)              # second derivative
for c in critical:
    label = "max" if f2.subs(x, c) < 0 else "min"
    print(f"x={c}: f''={f2.subs(x, c)} -> {label}")
# x=0: f''=-6 -> max
# x=2: f''=6  -> min

print("Inflection (f''=0):", sp.solve(f2, x))   # [1]
print("Exact x-intercepts:", sp.solve(f, x))    # [1, 1 - sqrt(3), 1 + sqrt(3)]

Computational Note. Notice sp.solve(f, x) returns the exact roots [1, 1 - sqrt(3), 1 + sqrt(3)], vindicating the §9.9 pitfall: the messy "$\approx 0.84$" decimal was simply wrong. Symbolic algebra refuses to round, so it catches the exact root $x = 1$ that a numerical solver alone might bury in floating-point noise. This is the fourth theme — hand computation builds understanding, machine computation builds power — in miniature: you find the structure, sympy confirms it without error.

The rational function's inflection points succumb just as easily:

import sympy as sp

x = sp.Symbol('x')
g = x / (x**2 + 1)
g2 = sp.diff(g, x, 2)
print("Inflection points:", sp.solve(g2, x))    # [0, -sqrt(3), sqrt(3)]

You can also let sympy resolve indeterminate limits directly — a check on your L'Hôpital work:

import sympy as sp

x = sp.Symbol('x')
print(sp.limit((x - sp.sin(x)) / x**3, x, 0))    # 1/6  (matches Example 3)
print(sp.limit(x**x, x, 0, '+'))                 # 1    (matches Example 5)

9.12 Applications Across Fields

The shape-analysis toolkit is field-agnostic. Here are two readings of the same calculus, reinforcing the theme that calculus appears in every quantitative field.

Pharmacokinetics (biology/medicine). After an intravenous dose, a drug's blood concentration often follows $C(t) = \frac{D}{V}\,e^{-kt}$, where $D$ is the dose, $V$ the volume of distribution, and $k > 0$ the elimination-rate constant. Then $C'(t) = -k\,C(t) < 0$: the concentration is always decreasing after the dose. And $C''(t) = -k\,C'(t) = k^2\,C(t) > 0$: the curve is always concave up — a steep initial drop that flattens out, never overshooting zero. The half-life $t_{1/2}$ solves $C(t_{1/2}) = \tfrac12 C(0)$, i.e. $e^{-kt_{1/2}} = \tfrac12$, giving $t_{1/2} = \frac{\ln 2}{k}$. Dosing schedules — "take every 8 hours" — are calibrated from exactly this derivative analysis.

Marginal analysis (economics). Economists define marginal cost as $C'(q)$, the derivative of total cost with respect to quantity $q$, and marginal revenue as $R'(q)$. Profit $\pi(q) = R(q) - C(q)$ is maximized where $\pi'(q) = R'(q) - C'(q) = 0$, i.e. marginal revenue equals marginal cost — the most cited rule in microeconomics. The second-order condition $\pi''(q) = R''(q) - C''(q) < 0$ (our second derivative test) confirms the critical point is a profit maximum rather than a minimum. Chapter 10 develops constrained optimization in full; here you see that the entire apparatus is just $f'$ and $f''$ wearing economic clothing.

A physics aside — projectile range. A projectile launched at angle $\theta$ with speed $v_0$ travels horizontal range $R(\theta) = \frac{v_0^2}{g}\sin(2\theta)$ (neglecting air resistance). Then $R'(\theta) = \frac{2v_0^2}{g}\cos(2\theta) = 0$ when $2\theta = \tfrac{\pi}{2}$, i.e. $\theta = 45^\circ$. Since $R''(\theta) = -\frac{4v_0^2}{g}\sin(2\theta) < 0$ there, it is a maximum: a $45^\circ$ launch maximizes range. One derivative recovers a fact known to every artillery officer for centuries.

Add to Your Modeling Portfolio. Add a shape analysis of your model's key function: find its critical points, classify them with the second derivative test, and identify intervals of concavity and any inflection points. Biology: analyze a drug-concentration curve $C(t) = \frac{D}{V}e^{-kt}$ — confirm it is decreasing and concave up, and compute its half-life $t_{1/2} = \frac{\ln 2}{k}$. Economics: find the production level where marginal revenue meets marginal cost, and use $\pi''(q) < 0$ to confirm it maximizes profit. Physics: for a potential energy function $U(x)$, locate equilibria as critical points and use $U''(x)$ to classify them as stable (minima) or unstable (maxima). Data Science: sketch a loss function $L(w)$ in one weight; identify its minimum and check convexity ($L'' > 0$) — the property that guarantees gradient descent (our anchor example from Chapter 6) converges to the global optimum.

9.13 Why This Matters

The derivative began as the slope of a tangent line. In this chapter it became something larger: a complete diagnostic instrument for the shape of a function. Critical points locate the peaks and valleys; the sign of $f'$ traces the climbs and descents; the sign of $f''$ reads the bends; inflection points mark the transitions. The Mean Value Theorem — which we proved from Rolle's theorem, which we proved from Fermat's theorem and the Extreme Value Theorem — is the rigorous foundation underneath all of it, and it will reappear as the keystone of the Fundamental Theorem of Calculus in Chapter 14.

These ideas radiate outward. Optimization (Chapter 10) is critical-point analysis applied to word problems. Linear approximation and Newton's method (Chapter 11) exploit concavity and tangent lines. In physics, equilibria are extrema of potential energy; in economics, optimal output is where marginals meet; in statistics, the mode of a distribution is a maximum and concavity controls its spread; in machine learning, the geometry of a loss surface — convex or not — decides whether optimization is easy or treacherous. Wherever a quantity is being maximized, minimized, or merely understood, the tools of this chapter are at work.

Looking Ahead

Chapter 10 dives into optimization: real-world problems that ask you to minimize cost, maximize volume, or find the shortest path. Every technique you built here — critical points, the closed-interval method, the second derivative test — becomes a workhorse there. Chapter 11 introduces linear approximation and Newton's method, using the tangent line (and the concavity you now understand) to approximate values and hunt down roots numerically. From there, Part II closes with antiderivatives (Chapter 12), and the great inversion of Chapter 14 — where the Mean Value Theorem you proved today returns to make the Fundamental Theorem of Calculus possible — comes into view.