Home / AP Calculus BC : 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points- Exam Style questions with Answer- MCQ

AP Calculus BC : 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points- Exam Style questions with Answer- MCQ

Question

The table above gives values of the continuous function f at selected values of x. If f has exactly two critical points on the open interval (10, 14), which of the following must be true?

\( x \)1011121314
\( f(x) \)52365

A) \( f(x) > 0 \) for all x in (10, 14)
B) \( f'(x) \) exists for all x in (10, 14)
C) \( f'(x) < 0 \) for all x in (10, 11)
D) \( f'(12) \neq 0 \)

▶️ Answer/Explanation
Solution

Step 1: Analyze critical points

Given f has exactly two critical points in (10,14), we can deduce:

  • The function must change direction twice (from decreasing to increasing or vice versa)
  • From the table, we see:
    • f decreases from x=10 to x=11 (5→2)
    • f increases from x=11 to x=13 (2→6)
    • f decreases from x=13 to x=14 (6→5)

Step 2: Evaluate options

A) \( f(x) > 0 \): Not necessarily true – we have no information about values between table points

B) \( f'(x) \) exists: Not necessarily true – critical points could be where derivative doesn’t exist

C) \( f'(x) < 0 \) on (10,11): While true for these discrete points, we can’t guarantee for all x in the interval

D) \( f'(12) \neq 0 \): Must be true because:

  • If f'(12)=0, this would be a third critical point (between the min at x≈11 and max at x≈13)
  • But we’re told there are exactly two critical points

Conclusion:

Only option D must be true given the conditions.

✅ Answer: D) \( f'(12) \neq 0 \)

Question

\(f(x)=1+x^{\frac{2}{3}}\) , which of the following is NOT true?

(A) f is continuous for all real numbers.
(B) f has a minimum at x = 0 .
(C) f is increasing for x > 0 .
(D) f ′( x) exists for all x.
(E) f ′′(x ) is negative for x > 0 .

▶️ Answer/Explanation
Solution

Step 1: Analyze the function properties

Given \( f(x) = 1 + x^{2/3} \), we examine each statement:

A) Continuity: True
\( x^{2/3} \) is defined and continuous for all real numbers.

B) Minimum at x=0: True
\( f(0) = 1 \), and \( f(x) \geq 1 \) for all x since \( x^{2/3} \geq 0 \).

C) Increasing for x>0: True
First derivative \( f'(x) = \frac{2}{3}x^{-1/3} > 0 \) for x > 0.

D) Differentiability: False
\( f'(x) = \frac{2}{3}x^{-1/3} \) is undefined at x=0 (vertical tangent).

E) Concavity: True
Second derivative \( f”(x) = -\frac{2}{9}x^{-4/3} < 0 \) for x > 0.

Step 2: Conclusion

Statement D is the one that is NOT true, as the derivative does not exist at x=0.

✅ Answer: D) f'(x) exists for all x

Question

 The graph of \(f ′\) , the derivative of f , is shown in the figure above. Which of the following describes all relative extrema of f on the open interval (a b,)?
(A) One relative maximum and two relative minima

(B) Two relative maxima and one relative minimum

(C) Three relative maxima and one relative minimum

(D) One relative maximum and three relative minima

(E) Three relative maxima and two relative minima

▶️ Answer/Explanation
Solution
Relative extrema of \( f \) occur where \( f’ = 0 \) and the sign of \( f’ \) changes. Analyze the graph of \( f’ \):
– \( f’ \) crosses the x-axis first (from negative to positive), indicating a relative minimum.
– \( f’ \) crosses the x-axis again (from positive to negative), indicating a relative maximum.
– \( f’ \) crosses the x-axis a third time (from negative to positive), indicating another relative minimum.
This gives one relative maximum and two relative minima.
✅ Answer: A)

Question

Let g be a continuous function on the closed interval [0,1]. Let (0)= 1 and g (1) =0 . Which of the following is NOT necessarily true?

(A) There exists a number h in [0,1] such that \(g(h)\geq g(x) \)for all x in [0,1].

(B) For all a and b in [0,1], if a =b , then g(a)=g(b).

(C) There exists a number h in [0,1] such that \(g(h)=\frac{1}{2}\)

(D) There exists a number h in [0,1] such that \(g(h)=\frac{3}{2}\)

(E) For all h in the open interval \((0,1),\lim_{x\rightarrow h}g(x)=g(h)\)

▶️ Answer/Explanation
Solution
A) True. By the Extreme Value Theorem, \( g \) attains a maximum in \([0, 1]\).
B) True. This is true by the definition of a function.
C) True. By the Intermediate Value Theorem, \( g \) takes all values between 1 and 0, including \( \frac{1}{2} \).
D) Not necessarily true. \( \frac{3}{2} > 1 \), and if \( g(x) = 1 – x \), \( g(x) \) ranges from 1 to 0, never reaching \( \frac{3}{2} \).
E) True. Continuity of \( g \) on \([0, 1]\) implies continuity at all \( h \) in \((0, 1)\).
✅ Answer: D)
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