Question

use the graph below to find exact values for the indicated derivatives, or state that they do not exist. if a derivative does not exist, enter dne in the answer blank. the graph of f(x) is black and has a sharp corner at x = 2. the graph of g(x) is blue. let v(x)=f(f(x)). find a. v(1)= b. v(2)= c. v(3)=

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 $y = f(x)$ for $x = 1$

For $x\in[0,2]$, the function $f(x)$ is a line with slope $m_1 = 2$ (using the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$ for two points on the line segment of $f(x)$ in this interval). For $x\in[2,4]$, the function $f(x)$ is a line with slope $m_2=-2$. When $x = 1$, $f(1)=2$, and $f^{\prime}(1)=2$. Then $v^{\prime}(1)=f^{\prime}(f(1))\cdot f^{\prime}(1)$. Since $f(1) = 2$, and $f^{\prime}(2)$ does not exist (due to the sharp corner at $x = 2$), but we consider the left - hand and right - hand derivatives. Since $f^{\prime}(1) = 2$ and $f^{\prime}(2)$ (left - hand derivative in the sub - domain where $x\in[0,2]$) is $2$, $v^{\prime}(1)=f^{\prime}(2)\cdot f^{\prime}(1)$. The left - hand derivative of $f(x)$ at $x = 2$ is $2$, so $v^{\prime}(1)=2\times2 = 4$.

Step3: Analyze the graph for $x = 2$

Since $f(x)$ has a sharp corner at $x = 2$, $f^{\prime}(2)$ does not exist. By the chain - rule $v^{\prime}(2)=f^{\prime}(f(2))\cdot f^{\prime}(2)$, and since $f^{\prime}(2)$ does not exist, $v^{\prime}(2)$ does not exist.

Step4: Analyze the graph for $x = 3$

When $x = 3$, $f(3)=1$, and $f^{\prime}(3)=-2$. Then $v^{\prime}(3)=f^{\prime}(f(3))\cdot f^{\prime}(3)$. Since $f(3)=1$ and $f^{\prime}(1)=2$, $v^{\prime}(3)=f^{\prime}(1)\cdot f^{\prime}(3)=2\times(-2)=-4$.

Answer:

A. $v^{\prime}(1)=4$
B. $v^{\prime}(2)=\text{dne}$
C. $v^{\prime}(3)=-4$