Question

the graphs of f (solid) and g (dashed) are given below. let (p(x)=f(x)cdot g(x)) and (q(x)=\frac{f(x)}{g(x)}) determine (p(4)). enter dne if the derivative fails to exist. determine (q(4)). enter dne if the derivative fails to exist. hint: take notice that p is the product of two functions whereas q is the quotient of two functions.

Explanation:

Step1: Recall product - rule

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

Step2: Find values from the graph at $x = 4$

From the graph, we first find $f(4)$, $g(4)$, $f^{\prime}(4)$ and $g^{\prime}(4)$.
$f(4)=0$, $g(4)=3$.
The slope of $f(x)$ at $x = 4$ (i.e., $f^{\prime}(4)$): The graph of $f(x)$ is a horizontal line for $x\geq1$, so $f^{\prime}(4)=0$.
The slope of $g(x)$ at $x = 4$ (i.e., $g^{\prime}(4)$): The graph of $g(x)$ is a straight - line for $x\geq3$. The line passes through $(3,0)$ and $(5,3)$. Using the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$, we have $g^{\prime}(4)=\frac{3-0}{5 - 3}=\frac{3}{2}$.

Step3: Calculate $p^{\prime}(4)$

Substitute into the product - rule formula: $p^{\prime}(4)=f^{\prime}(4)g(4)+f(4)g^{\prime}(4)$.
$p^{\prime}(4)=(0)\times(3)+(0)\times\frac{3}{2}=0$.

Step4: Recall quotient - rule

The quotient - rule states that if $q(x)=\frac{f(x)}{g(x)}$, then $q^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$.

Step5: Calculate $q^{\prime}(4)$

Substitute $f(4) = 0$, $g(4)=3$, $f^{\prime}(4)=0$ and $g^{\prime}(4)=\frac{3}{2}$ into the quotient - rule formula.
$q^{\prime}(4)=\frac{(0)\times(3)-(0)\times\frac{3}{2}}{3^{2}}=0$.

Answer:

$p^{\prime}(4)=0$
$q^{\prime}(4)=0$