AP Calculus BC 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema - FRQs - Exam Style Questions

(a) Find \(f(-6)\) and \(f(5)\)
By the Fundamental Theorem of Calculus, \[ f(b)=f(a)+\int_a^b f'(x)\,dx. \] • From \(-6\) to \(-2\), \(f'(x)\) is a line segment dropping from \(2\) to \(0\). Area \(=\) trapezoid \(=\frac{(2+0)}{2}\cdot 4 = 4\). Hence \[ f(-6)=f(-2)+\int_{-2}^{-6}\!f'(x)\,dx=7-4=\boxed{3}. \] • From \(-2\) to \(2\), \(f'(x)\) is a semicircle of radius \(2\) below the axis → area \(=-\frac12\pi(2)^2=-2\pi\). From \(2\) to \(5\), the line segment above the axis contributes area \(+3\). Therefore \[ f(5)=f(-2)+\int_{-2}^{5}\!f'(x)\,dx=7+(-2\pi)+3=\boxed{10-2\pi}. \]
(b) Intervals where \(f\) is increasing
\(f\) increases when \(f'(x)>0\). From the graph, \(f'(x)>0\) on \(\boxed{[-6,-2)}\) and \(\boxed{(2,5]}\).
(c) Absolute minimum of \(f\) on \([-6,5]\)
Check endpoints and any critical points where \(f'(x)=0\). From the figure, zeros at \(x=-2\) and \(x=2\). Values (from part (a) computations): \[ \begin{array}{c|c} x & f(x) \\ \hline -6 & 3 \\ -2 & 7 \\ 2 & 7-2\pi \\ 5 & 10-2\pi \end{array} \] The least value is at \(x=2\): \(\boxed{f_{\min}=7-2\pi}\).
(d) \(f”(-5)\) and \(f”(3)\)
• On the linear segment from \((-6,2)\) to \((-2,0)\), the slope of \(f’\) is \[ f”(x)=\frac{\Delta f’}{\Delta x}=\frac{0-2}{-2-(-6)}=-\frac{1}{2}. \] Hence \(\boxed{f”(-5)=-\tfrac12}\).

• At \(x=3\), the graph of \(f’\) has a sharp peak (corner). One-sided slopes of \(f’\) differ: \[ \lim_{x\to 3^-}\frac{f'(x)-f'(3)}{x-3}=2,\qquad \lim_{x\to 3^+}\frac{f'(x)-f'(3)}{x-3}=-1. \] Since these are not equal, \(f”(3)\) \(\boxed{\text{does not exist}}\).