Review Extreme Value Theorem problems by testing endpoint and critical-point values on closed intervals and comparing candidates to justify the absolute result.
The graph of the differentiable function f, shown for −6≤x≤7, has a horizontal tangent at x=-2 and is linear for 0≤x≤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≤x≤6 at which the graph of g has a critical point. Give a reason for your answer.
g′(x)=f(x)
Fundamental
Theorem of Calculus
1 point
g′(x)=f(x)=0⇒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′=f in this part.
○ A response that writes g′′=f′ 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′(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