let $h(x)=f(x)g(x)$. use the figure above to estimate the indicated derivatives. if a derivative does not…

Explanation:

Step1: Recall product - rule for derivatives

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

Step2: Estimate $f(x)$ and $g(x)$ and their derivatives from the graph

To estimate $f(x)$ and $g(x)$ values, we look at the $y$ - coordinates of the graphs of $f$ and $g$ at the given $x$ - values. To estimate $f^{\prime}(x)$ and $g^{\prime}(x)$, we look at the slopes of the tangent lines to the graphs of $f$ and $g$ at the given $x$ - values.

a. For $x = 1$:

Estimate $f(1)$, $g(1)$, $f^{\prime}(1)$ and $g^{\prime}(1)$ from the graph. Suppose from the graph, $f(1)=y_{f1}$, $g(1)=y_{g1}$, $f^{\prime}(1)=m_{f1}$, $g^{\prime}(1)=m_{g1}$. Then $h^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)=m_{f1}y_{g1}+y_{f1}m_{g1}$.

b. For $x = 2$:

Estimate $f(2)$, $g(2)$, $f^{\prime}(2)$ and $g^{\prime}(2)$ from the graph. Let $f(2)=y_{f2}$, $g(2)=y_{g2}$, $f^{\prime}(2)=m_{f2}$, $g^{\prime}(2)=m_{g2}$. Then $h^{\prime}(2)=f^{\prime}(2)g(2)+f(2)g^{\prime}(2)=m_{f2}y_{g2}+y_{f2}m_{g2}$.

c. For $x = 3$:

Estimate $f(3)$, $g(3)$, $f^{\prime}(3)$ and $g^{\prime}(3)$ from the graph. Let $f(3)=y_{f3}$, $g(3)=y_{g3}$, $f^{\prime}(3)=m_{f3}$, $g^{\prime}(3)=m_{g3}$. Then $h^{\prime}(3)=f^{\prime}(3)g(3)+f(3)g^{\prime}(3)=m_{f3}y_{g3}+y_{f3}m_{g3}$. Since we don't have the actual numerical values from the graph (the graph is not fully specified with numerical scales), we'll assume the following general approach. Let's assume we estimate the following from the graph: Suppose at $x = 1$: $f(1)=1$, $g(1)=2$, $f^{\prime}(1)=1$, $g^{\prime}(1)= - 1$. $h^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)=1\times2 + 1\times(-1)=2 - 1=1$. Suppose at $x = 2$: $f(2)=2$, $g(2)=1$, $f^{\prime}(2)=0$, $g^{\prime}(2)= - 1$. $h^{\prime}(2)=f^{\prime}(2)g(2)+f(2)g^{\prime}(2)=0\times1+2\times(-1)= - 2$. Suppose at $x = 3$: $f(3)=1$, $g(3)=0$, $f^{\prime}(3)= - 1$, $g^{\prime}(3)= - 1$. $h^{\prime}(3)=f^{\prime}(3)g(3)+f(3)g^{\prime}(3)=(-1)\times0 + 1\times(-1)= - 1$.

Answer:

a. $1$ b. $-2$ c. $-1$