use the graph below to find exact values for the indicated derivatives, or state that they do not exist. if…

Explanation:

Step1: Recall chain - rule

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

Step2: Analyze the graph of $y = f(x)$ to find slopes

For $x\in[0,2]$, the equation of the line of $y = f(x)$ is $y = x$ (using the two - point form with $(0,0)$ and $(2,2)$), so $f^{\prime}(x)=1$ for $x\in[0,2)$. For $x\in[2,4]$, the equation of the line of $y = f(x)$ is $y=-x + 4$ (using the two - point form with $(2,2)$ and $(4,0)$), so $f^{\prime}(x)=-1$ for $x\in(2,4]$.

Step3: Find $v^{\prime}(1)$

First, when $x = 1$, $f(1)=1$ and $f^{\prime}(1)=1$. Then, using the chain - rule $v^{\prime}(1)=f^{\prime}(f(1))\cdot f^{\prime}(1)$. Since $f(1)=1$ and $f^{\prime}(1)=1$, and $f^{\prime}(f(1))=f^{\prime}(1)=1$, we have $v^{\prime}(1)=1\times1 = 1$.

Step4: Find $v^{\prime}(3)$

When $x = 3$, $f(3)=-3 + 4=1$. And $f^{\prime}(3)=-1$, $f^{\prime}(f(3))=f^{\prime}(1)=1$. Using the chain - rule $v^{\prime}(3)=f^{\prime}(f(3))\cdot f^{\prime}(3)=1\times(-1)=-1$.

Answer:

A. $v^{\prime}(1)=1$ B. $v^{\prime}(2)=DNE$ C. $v^{\prime}(3)=-1$