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 $y = g(x)$ (derivatives)

The slope of a linear function $y = mx + b$ is $m$. For a linear function passing through two points $(x_1,y_1)$ and $(x_2,y_2)$, the slope $m=\frac{y_2 - y_1}{x_2 - x_1}$.
For $y = f(x)$: Let's take two points, say $(0,2)$ and $(4,6)$. Then $f^{\prime}(x)=\frac{6 - 2}{4-0}=1$. So, $f^{\prime}(4)=1$.
For $y = g(x)$: Let's take two points, say $(0,3)$ and $(4,5)$. Then $g^{\prime}(x)=\frac{5 - 3}{4 - 0}=\frac{1}{2}$. So, $g^{\prime}(4)=\frac{1}{2}$.

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

From the graph of $y = f(x)$, when $x = 4$, $y=f(4)=6$. From the graph of $y = g(x)$, when $x = 4$, $y = g(4)=5$.

Step4: Substitute values into the product - rule formula

$h^{\prime}(4)=f^{\prime}(4)g(4)+f(4)g^{\prime}(4)$. Substitute $f^{\prime}(4)=1$, $g(4)=5$, $f(4)=6$, and $g^{\prime}(4)=\frac{1}{2}$ into the formula.
$h^{\prime}(4)=1\times5+6\times\frac{1}{2}=5 + 3=8$.

Answer:

$8$