Question

let ( h(x) = f(x) cdot g(x) ), and ( k(x) = \frac{f(x)}{g(x)} ). use the figures below to find the values of the indicated derivatives. note: when estimating values from the graphs, each is a multiple of ( 0.5 ) (e.g. ( -1, -0.5, 0, 0.5, ) etc.). (enter dne for any answer where the derivative does not exist.) a. ( h(-3) = ) b. ( k(3) = )

Explanation:

Step1: Recall product rule for $h'(x)$

$h(x)=f(x)g(x)$, so $h'(x)=f'(x)g(x)+f(x)g'(x)$

Step2: Read values at $x=-3$

From graphs: $f(-3)=-3$, $g(-3)=-4$, $f'(-3)=\frac{-2-(-4)}{-2-(-3)}=2$, $g'(-3)=\frac{-8-(-4)}{-2-(-3)}=-4$

Step3: Calculate $h'(-3)$

$h'(-3)=(2)(-4)+(-3)(-4)=-8+12=4$

Step4: Recall quotient rule for $k'(x)$

$k(x)=\frac{f(x)}{g(x)}$, so $k'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$

Step5: Read values at $x=3$

From graphs: $f(3)=-4$, $g(3)=-4$, $f'(3)=\frac{-2-0}{2-3}=2$, $g'(3)=\frac{-8-(-4)}{2-3}=4$

Step6: Calculate $k'(3)$

$k'(3)=\frac{(2)(-4)-(-4)(4)}{(-4)^2}=\frac{-8+16}{16}=\frac{8}{16}=\frac{1}{2}$

Answer:

A. $h'(-3)=4$
B. $k'(3)=\frac{1}{2}$