AP Calculus BC 5.2: Extreme Value Theorem

Practice AP Calculus BC questions on using continuity and critical-point conditions to justify extrema on closed intervals.

[Maximum number: 2]

The graph of the differentiable function f, shown for $-6 \leq x \leq 7$ , has a horizontal tangent at x=-2 and is linear for $0 \leq x \leq 7$ . Let R be the region in the second quadrant bounded by the graph of f, the vertical line x=-6, and the x - and y-axes. Region R has area 12.

For the function g defined in part (a), find all values of x in the interval $0 \leq x \leq 6$ at which the graph of g has a critical point. Give a reason for your answer.

$g^{\prime}(x)=f(x)$

Fundamental
Theorem of Calculus

1 point

$g^{\prime}(x)=f(x)=0 \Rightarrow x=4$

Answer with reason

1 point

Therefore, the graph of g has a critical point at x=4.

Scoring notes:

- The first point is earned for explicitly making the connection $g^{\prime}=f$ in this part.

○ A response that writes $g^{\prime \prime}=f^{\prime}$ earns the first point but can only earn the second point by

reasoning from f=0.

- A response that does not earn the first point is eligible to earn the second point with an implied

application of the FTC (e.g., "Because $g^{\prime}(4)=0, x=4$ is a critical point").

- A response that reports any additional critical points in 0<x<6 does not earn the second point.

○ Any presented critical point outside the interval 0<x<6 will not affect scoring.

Total for part (b) 2 points

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FUN-1.C—Justify conclusions about functions by applying the Extreme Value Theorem question 1