use the figure to find the indicated derivatives, if they exist. (if an answer does not exist, enter dne.)…

Explanation:

Step1: Recall sum - rule of derivatives

The sum - rule of derivatives states that if $h(x)=f(x)+g(x)$, then $h^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)$.

Step2: Find the slope of $f(x)$ at $x = 2$

The function $f(x)$ is a straight - line. The equation of a line is $y=mx + b$, where $m$ is the slope. For $f(x)$, two points on the line are $(0,4)$ and $(4,2)$. Using the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$, we have $m_f=\frac{2 - 4}{4 - 0}=-\frac{1}{2}$. So $f^{\prime}(x)=-\frac{1}{2}$ for all $x$ in the domain of $f(x)$.

Step3: Find the slope of $g(x)$ at $x = 2$

The function $g(x)$ is composed of two line - segments. For $x\in[0,3]$, the line of $g(x)$ passes through $(0,0)$ and $(3,1)$. Using the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$, we get $m_g=\frac{1 - 0}{3 - 0}=\frac{1}{3}$. So $g^{\prime}(x)=\frac{1}{3}$ for $x\in(0,3)$.

Step4: Calculate $h^{\prime}(2)$

Since $h^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)$, when $x = 2$, $h^{\prime}(2)=f^{\prime}(2)+g^{\prime}(2)$. Substituting $f^{\prime}(2)=-\frac{1}{2}$ and $g^{\prime}(2)=\frac{1}{3}$ into the formula, we have $h^{\prime}(2)=-\frac{1}{2}+\frac{1}{3}=-\frac{3 - 2}{6}=-\frac{1}{6}$.

Step5: Analyze the derivative of $g(x)$ at $x = 3$

The function $g(x)$ has a sharp corner at $x = 3$. The left - hand derivative of $g(x)$ at $x = 3$ is $\frac{1}{3}$ (from the line segment for $x\in[0,3]$) and the right - hand derivative of $g(x)$ at $x = 3$ is found using the two points $(3,1)$ and $(5,3)$. The slope of the right - hand line segment is $m=\frac{3 - 1}{5 - 3}=1$. Since the left - hand derivative and the right - hand derivative of $g(x)$ at $x = 3$ are not equal, $g^{\prime}(3)$ does not exist. Since $h^{\prime}(3)=f^{\prime}(3)+g^{\prime}(3)$ and $g^{\prime}(3)$ does not exist, $h^{\prime}(3)$ does not exist (DNE).

Answer:

(a) $-\frac{1}{6}$ (b) DNE