Question

the graphs of the linear function $f$ and the piece - wise linear function $g$ are shown in the figure above. if $h(x)=f(x)g(x)$, then $h(3)=$

Explanation:

Step1: Recall product - rule of differentiation

The product - rule states that if $h(x)=f(x)g(x)$, then $h^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. So, $h^{\prime}(3)=f^{\prime}(3)g(3)+f(3)g^{\prime}(3)$.

Step2: Find $f(3)$ and $g(3)$ from the graph

From the graph, when $x = 3$, $f(3)=4$ and $g(3)=3$.

Step3: Find the slope of $y = f(x)$ (i.e., $f^{\prime}(x)$)

The function $y = f(x)$ is a straight - line. The slope of a line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $y = f(x)$, using the points $(0,7)$ and $(7,0)$, the slope $f^{\prime}(x)=\frac{0 - 7}{7 - 0}=- 1$. So, $f^{\prime}(3)=-1$.

Step4: Find the slope of $y = g(x)$ (i.e., $g^{\prime}(x)$) at $x = 3$

For $y = g(x)$ near $x = 3$, using the points $(2,1)$ and $(4,5)$, the slope $g^{\prime}(x)=\frac{5 - 1}{4 - 2}=2$. So, $g^{\prime}(3)=2$.

Step5: Calculate $h^{\prime}(3)$

Substitute $f(3)=4$, $g(3)=3$, $f^{\prime}(3)=-1$, and $g^{\prime}(3)=2$ into the product - rule formula $h^{\prime}(3)=f^{\prime}(3)g(3)+f(3)g^{\prime}(3)$.
$h^{\prime}(3)=(-1)\times3+4\times2=-3 + 8=5$.

Answer:

$5$