use the figures below to evaluate the indicated derivative, or state that it does not exist. if the…

Explanation:

Step1: Recall chain - rule

The chain - rule states that $\frac{d}{dx}f(g(x))=f^{\prime}(g(x))\cdot g^{\prime}(x)$. We need to find $g(100)$ and $g^{\prime}(100)$ and $f^{\prime}(g(100))$.

Step2: Find $g^{\prime}(x)$ from the graph of $g(x)$

The graph of $g(x)$ is a non - vertical straight line for $x < 120$. The slope of a straight - line $y = mx + b$ is its derivative. For $g(x)$, using two points on the line of $g(x)$ (say $(0, y_1)$ and $(120,y_2)$), the slope $g^{\prime}(x)$ is constant. Let's assume the line of $g(x)$ passes through $(0,40)$ and $(120,160)$. Then $g^{\prime}(x)=\frac{160 - 40}{120-0}=\frac{120}{120}=1$ for $x<120$. Since $100<120$, $g^{\prime}(100) = 1$.

Step3: Find $g(100)$

Using the point - slope form of a line $y - y_1=m(x - x_1)$ with $m = 1$, $x_1 = 0$ and $y_1 = 40$, we have $g(x)=x + 40$. So $g(100)=100 + 40=140$.

Step4: Find $f^{\prime}(g(100))=f^{\prime}(140)$

The graph of $f(x)$ is composed of line segments. For $x>80$, the line segment of $f(x)$ has a slope. Using two points on the right - hand line segment of $f(x)$ (say $(80,0)$ and $(160,80)$), the slope of this line segment is $f^{\prime}(x)=\frac{80 - 0}{160 - 80}=\frac{80}{80}=1$ for $x>80$. Since $140>80$, $f^{\prime}(140)=1$.

Step5: Calculate $\frac{d}{dx}f(g(x))|_{x = 100}$

By the chain - rule $\frac{d}{dx}f(g(x))|{x = 100}=f^{\prime}(g(100))\cdot g^{\prime}(100)$. Substituting $f^{\prime}(g(100))=f^{\prime}(140) = 1$ and $g^{\prime}(100)=1$, we get $\frac{d}{dx}f(g(x))|{x = 100}=1\times1 = 1$.

Answer:

$1$