Question

the graphs of f (solid) and g (dashed) are given below. let (h(x)=f(g(x))) and (k(x)=g(f(x))). determine (h(5)). enter dne if the derivative fails to exist. determine (k(5)). enter dne if the derivative fails to exist. hint: take notice that both h and k are composite functions.

Explanation:

Step1: Recall the chain - rule

The chain - rule states that if $h(x)=f(g(x))$, then $h^{\prime}(x)=f^{\prime}(g(x))\cdot g^{\prime}(x)$. And if $k(x)=g(f(x))$, then $k^{\prime}(x)=g^{\prime}(f(x))\cdot f^{\prime}(x)$.

Step2: Find $h^{\prime}(5)$

First, find $g(5)$ from the graph. From the graph of $g(x)$ (dashed line), when $x = 5$, $g(5)=0$. Then, $h^{\prime}(5)=f^{\prime}(g(5))\cdot g^{\prime}(5)=f^{\prime}(0)\cdot g^{\prime}(5)$.
The slope of $g(x)$ at $x = 5$: The graph of $g(x)$ is a straight - line segment for $x\geq3$ with two points $(3,0)$ and $(6,-6)$. The slope $g^{\prime}(5)=\frac{-6 - 0}{6 - 3}=-2$.
The slope of $f(x)$ at $x = 0$: The graph of $f(x)$ (solid line) is a straight - line segment for $x\geq - 1$ with two points $(-1,1)$ and $(1,-2)$. The slope $f^{\prime}(0)=\frac{-2 - 1}{1+1}=-\frac{3}{2}$.
So, $h^{\prime}(5)=f^{\prime}(0)\cdot g^{\prime}(5)=(-\frac{3}{2})\times(-2)=3$.

Step3: Find $k^{\prime}(5)$

First, find $f(5)$ from the graph. From the graph of $f(x)$ (solid line), when $x = 5$, $f(5)=3$. Then, $k^{\prime}(5)=g^{\prime}(f(5))\cdot f^{\prime}(5)=g^{\prime}(3)\cdot f^{\prime}(5)$.
The slope of $f(x)$ at $x = 5$: The graph of $f(x)$ is a straight - line segment for $x\geq - 1$ with two points $(-1,1)$ and $(1,-2)$ and then another segment for $x\geq1$. For $x\geq1$, the segment has two points $(1,-2)$ and $(5,3)$. The slope $f^{\prime}(5)=\frac{3 + 2}{5 - 1}=\frac{5}{4}$.
The slope of $g(x)$ at $x = 3$: The graph of $g(x)$ has a sharp - point at $x = 3$, so the derivative $g^{\prime}(3)$ does not exist. Thus, $k^{\prime}(5)$ does not exist, and we write $k^{\prime}(5)=\text{DNE}$.

Answer:

$h^{\prime}(5)=3$
$k^{\prime}(5)=\text{DNE}$