AP Calculus AB 6.4 The Fundamental Theorem of Calculus and Accumulation Functions - FRQs - Exam Style Questions

(A) \(g'(8)\)
By the Fundamental Theorem of Calculus (accumulation functions), \(g'(x)=f(x)\). Therefore \(g'(8)=f(8)\).
On \([6,12]\), \(f\) is the line through \((6,0)\) and \((12,3)\) with slope \(\dfrac{3-0}{12-6}=\dfrac{1}{2}\). Its equation is \(f(x)=\dfrac{1}{2}(x-6)\).
Hence \(g'(8)=f(8)=\dfrac{1}{2}\times(8-6)=\boxed{1}.\)

(B) Points of inflection of \(g\)
Since \(g'(x)=f(x)\), we have \(g”(x)=f'(x)\). A point of inflection of \(g\) occurs where \(g”\) changes sign, i.e., where \(f’\) changes sign.
From the graph of \(f\):
• At \(x=-3\) (bottom of the left semicircle), \(f\) changes from decreasing to increasing \(\Rightarrow f’\) changes sign \(\Rightarrow\) \(g\) has an inflection at \(x=-3\).
• At \(x=3\) (top of the right semicircle), \(f\) changes from increasing to decreasing \(\Rightarrow f’\) changes sign \(\Rightarrow\) \(g\) has an inflection at \(x=3\).
• At \(x=6\) (join from semicircle to line), the slope of \(f\) jumps from negative (left) to positive (right) \(\Rightarrow f’\) changes sign \(\Rightarrow\) \(g\) has an inflection at \(x=6\).
Therefore the inflection \(x\)-values are \(\boxed{x=-3,\,3,\,6}\).

(C) \(g(12)\) and \(g(0)\)
By definition, \(g(12)=\displaystyle\int_{6}^{12} f(t)\,dt\). On \([6,12]\), the region is a triangle above the \(x\)-axis with base \(6\) and height \(3\).
Thus \(g(12)=\dfrac{1}{2}\times 6\times 3=\boxed{9}.\)
Next, \(g(0)=\displaystyle\int_{6}^{0} f(t)\,dt=-\int_{0}^{6} f(t)\,dt\). On \([0,6]\), the region is a semicircle of radius \(3\) above the \(x\)-axis, so the area is \(\dfrac{1}{2}\pi\times 3^{2}=\dfrac{9\pi}{2}\).
Therefore \(g(0)=-\dfrac{9\pi}{2}=\boxed{-\dfrac{9\pi}{2}}.\)

(D) Absolute minimum of \(g\) on \([-6,12]\)
Candidates Test: absolute extrema occur at critical points (\(g'(x)=0\)) or endpoints \(x=-6,12\). Since \(g'(x)=f(x)\), solve \(f(x)=0\). From the graph, zeros of \(f\) in \([-6,12]\) are \(x=-6,0,6\).
Evaluate \(g\) at candidates:
• \(g(-6)=\displaystyle\int_{6}^{-6}\!f=\,-\!\int_{-6}^{6}\!f\). The area on \([-6,0]\) is a semicircle of radius \(3\) below the axis (value \(-\dfrac{9\pi}{2}\)), and on \([0,6]\) is the same area above the axis (\(+\dfrac{9\pi}{2}\)). They cancel, so \(g(-6)=0\).
• \(g(0)=\boxed{-\dfrac{9\pi}{2}}\) (from part (C)).
• \(g(6)=\displaystyle\int_{6}^{6}\!f=0\).
• \(g(12)=\boxed{9}\) (from part (C)).
Comparing values: \(0,\ -\dfrac{9\pi}{2},\ 0,\ 9\). The smallest is \(-\dfrac{9\pi}{2}\) at \(x=0\).
Therefore, \(g\) attains its absolute minimum at \(\boxed{x=0}\).