Question

question 7
if $f(x)=g(x)cdot h(x)$, then
$f(1)=$
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Explanation:

Step1: Recall product - rule

The product - rule states that if $f(x)=g(x)\cdot h(x)$, then $f^{\prime}(x)=g^{\prime}(x)h(x)+g(x)h^{\prime}(x)$. So, $f^{\prime}(1)=g^{\prime}(1)h(1)+g(1)h^{\prime}(1)$.

Step2: Find $g(1)$ and $g^{\prime}(1)$ from the graph of $g(x)$

The graph of $g(x)$ is a straight - line passing through the points $(0,0)$ and $(5,5)$. The equation of the line $g(x)$ using the slope - intercept form $y = mx + b$ (where $b = 0$ and $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{5 - 0}{5 - 0}=1$), so $g(x)=x$. Then $g(1)=1$ and $g^{\prime}(x) = 1$, so $g^{\prime}(1)=1$.

Step3: Find $h(1)$ and $h^{\prime}(1)$ from the graph of $h(x)$

From the graph of $h(x)$, when $x = 1$, $h(1)=1$. The graph of $h(x)$ for $x\in[0,3]$ is a straight - line with slope $m=\frac{5 - 1}{3 - 1}=\frac{4}{2}=2$. So, $h^{\prime}(1)=2$.

Step4: Calculate $f^{\prime}(1)$

Substitute $g(1)=1$, $g^{\prime}(1)=1$, $h(1)=1$, and $h^{\prime}(1)=2$ into the product - rule formula $f^{\prime}(1)=g^{\prime}(1)h(1)+g(1)h^{\prime}(1)$. We get $f^{\prime}(1)=1\times1 + 1\times2=3$.

Answer:

$3$