Chapter 4 — Quiz

10 questions, ~20 minutes. Aim for 7/10. Each answer cites the section to review.

1. A function $f$ is continuous at $a$ when:

  • A) $f(a)$ is defined
  • B) $\lim_{x \to a} f(x)$ exists
  • C) $f(a)$ is defined, $\lim_{x \to a} f(x)$ exists, and the two are equal
  • D) $f$ has a graph somewhere without gaps
Answer**C)** All three conditions must hold simultaneously; any one can fail while the others hold. *Reference: Section 4.2.*

2. $f(x) = \dfrac{x^2 - 9}{x - 3}$ has a discontinuity at $x = 3$ that is:

  • A) Removable
  • B) Jump
  • C) Infinite
  • D) Oscillating
Answer**A) Removable.** Factor and cancel: $f(x) = x + 3$ for $x \neq 3$, so $\lim_{x\to 3} f(x) = 6$ exists but $f(3)$ is undefined. Defining $f(3) = 6$ removes the hole. *Reference: Section 4.3.*

3. The Intermediate Value Theorem requires that the function be:

  • A) Continuous on a closed interval $[a,b]$
  • B) Differentiable on the interval
  • C) Bounded but possibly discontinuous
  • D) Monotonic
Answer**A)** IVT requires only continuity on the closed interval $[a, b]$ — no differentiability or monotonicity. *Reference: Section 4.6.*

4. Which function is not continuous at $x = 0$?

  • A) $f(x) = x^3$
  • B) $f(x) = e^x$
  • C) $f(x) = 1/x$
  • D) $f(x) = \sin x$
Answer**C)** $1/x$ is undefined at $0$ (condition 1 fails) and in fact has an infinite discontinuity there. *Reference: Sections 4.2, 4.3.*

5. Does $f(x) = x^3 - 4x + 1$ have a root in $[0, 2]$, and does IVT prove it?

  • A) Yes — IVT applied to $[0,2]$ directly, since $f(0)$ and $f(2)$ have opposite signs
  • B) Yes — IVT proves a root in $[0,1]$, since $f(0) = 1 > 0$ and $f(1) = -2 < 0$
  • C) No root exists in $[0,2]$
  • D) A root exists, but IVT cannot detect it
Answer**B)** At the endpoints $f(0) = 1$ and $f(2) = 1$ — *same* sign, so IVT on $[0,2]$ directly is inconclusive. But $f(1) = -2 < 0$, so on $[0,1]$ the signs differ ($1 > 0 > -2$) and IVT guarantees a root in $(0,1)$. (Option A is wrong because the endpoint signs match.) *Reference: Section 4.6.*

6. The bisection method converges at what rate?

  • A) Linear — it halves the bracketing interval each step (about one bit of precision per step)
  • B) Quadratic — it doubles the number of correct digits each step
  • C) Cubic
  • D) It is not guaranteed to converge
Answer**A) Linear.** Each step halves the interval, gaining one bit (~0.3 decimal digit) per iteration; convergence is guaranteed whenever the endpoints bracket a sign change. (Quadratic convergence describes Newton's method, Chapter 11.) *Reference: Section 4.7.*

7. For $f(x) = \dfrac{\sin x}{x}$ (undefined at $0$), redefining $f(0) = 1$ makes $f$:

  • A) Continuous everywhere
  • B) Still discontinuous at $0$
  • C) Continuous but the discontinuity becomes a jump
  • D) Discontinuous at every multiple of $\pi$
Answer**A) Continuous everywhere.** $\lim_{x \to 0} \frac{\sin x}{x} = 1$, so setting $f(0) = 1$ matches the limit and removes the (removable) discontinuity. *Reference: Sections 4.2, 4.3.*

8. The Extreme Value Theorem guarantees that a function continuous on a closed, bounded interval $[a, b]$:

  • A) Has a derivative everywhere
  • B) Attains both its absolute maximum and its absolute minimum
  • C) Is monotone
  • D) Has no roots
Answer**B)** EVT: a continuous function on a closed *and* bounded interval attains (achieves, not merely approaches) both its absolute max and min. *Reference: Section 4.9.*

9. Which type of discontinuity cannot be removed by redefining the function at a single point?

  • A) Removable
  • B) Jump
  • C) A hole in the graph
  • D) Both A and C
Answer**B) Jump.** A removable discontinuity (a hole, or a single misplaced value) is fixed by one redefinition. A jump cannot be: no single value can equal two different one-sided limits at once. *Reference: Section 4.3.*

10. Which is a function continuous on the open interval $(0, 1)$ but not on the closed interval $[0, 1]$?

  • A) No such function can exist
  • B) $f(x) = 1/x$
  • C) $f(x) = \sin(1/x)$
  • D) Both B and C
Answer**D)** Both $1/x$ (infinite discontinuity) and $\sin(1/x)$ (oscillating discontinuity) are continuous throughout $(0,1)$ but fail at the endpoint $x = 0$, so neither is continuous on $[0,1]$. *Reference: Sections 4.3, 4.8.*

Scoring

  • 9–10: Excellent — you own the three-condition definition, the discontinuity zoo, and both big theorems.
  • 7–8: Solid. Review the IVT and bisection material (Sections 4.6–4.7).
  • 5–6: Re-read the chapter and redo Exercises Parts B and D.
  • Below 5: Slow down and rework Sections 4.2–4.3 from scratch. Continuity is the load-bearing wall behind every later chapter.