Question

the graphs of the linear function $f$ and the linear function $g$ are shown in the figure above. if $h(x)=f(x)g(x)$, then $h(4)=$

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, $h^{\prime}(4)=f^{\prime}(4)g(4)+f(4)g^{\prime}(4)$.

Step2: Find the slope of $y = f(x)$ and $y = g(x)$

For a linear function $y = mx + b$, the slope $m$ can be found using the formula $m=\frac{y_2 - y_1}{x_2 - x_1}$.
For $y = f(x)$: Two points on the line are $(2,0)$ and $(6,4)$. Then $f^{\prime}(x)=\frac{4 - 0}{6 - 2}=1$.
For $y = g(x)$: Two points on the line are $(3,0)$ and $(6,4)$. Then $g^{\prime}(x)=\frac{4 - 0}{6 - 3}=\frac{4}{3}$.

Step3: Find $f(4)$ and $g(4)$ from the graph

From the graph of $y = f(x)$, when $x = 4$, $f(4)=2$.
From the graph of $y = g(x)$, when $x = 4$, $g(4)=\frac{4}{3}$.

Step4: Calculate $h^{\prime}(4)$

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

Answer:

$4$