Question

the graphs of the functions f and g are shown in the figure. let u(x)=f(x)g(x) and v(x)=\frac{f(x)}{g(x)}. (a) find u(1). (b) find v(5).

Explanation:

Step1: Recall product - rule for $u(x)$

The product - rule states that if $u(x)=f(x)g(x)$, then $u^{\prime}(x)=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$. To find $u^{\prime}(1)$, we need to find $f(1)$, $f^{\prime}(1)$, $g(1)$ and $g^{\prime}(1)$ from the graph.
From the graph, $f(1) = 0$, $g(1)=2$.
To find the slopes $f^{\prime}(1)$ and $g^{\prime}(1)$:
For $y = f(x)$ near $x = 1$, the function $f(x)$ is a line segment. The line passes through $(0,0)$ and $(2,4)$. The slope $m_f=\frac{4 - 0}{2-0}=2$, so $f^{\prime}(1)=2$.
For $y = g(x)$ near $x = 1$, the function $g(x)$ is a line segment. The line passes through $(0,2)$ and $(2,0)$. The slope $m_g=\frac{0 - 2}{2 - 0}=-1$, so $g^{\prime}(1)=-1$.
Substitute into the product - rule formula: $u^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)$.
$u^{\prime}(1)=(2)\times(2)+(0)\times(-1)=4$.

Step2: Recall quotient - rule for $v(x)$

The quotient - rule states that if $v(x)=\frac{f(x)}{g(x)}$, then $v^{\prime}(x)=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}$. To find $v^{\prime}(5)$, we need to find $f(5)$, $f^{\prime}(5)$, $g(5)$ and $g^{\prime}(5)$ from the graph.
From the graph, $f(5)=3$, $g(5)=2$.
For $y = f(x)$ near $x = 5$, the function $f(x)$ is a line segment. The line passes through $(2,4)$ and $(6,2)$. The slope $m_f=\frac{2 - 4}{6 - 2}=-\frac{1}{2}$, so $f^{\prime}(5)=-\frac{1}{2}$.
For $y = g(x)$ near $x = 5$, the function $g(x)$ is a line segment. The line passes through $(2,0)$ and $(6,3)$. The slope $m_g=\frac{3 - 0}{6 - 2}=\frac{3}{4}$, so $g^{\prime}(5)=\frac{3}{4}$.
Substitute into the quotient - rule formula:
$v^{\prime}(5)=\frac{f^{\prime}(5)g(5)-f(5)g^{\prime}(5)}{g^{2}(5)}$.
$v^{\prime}(5)=\frac{(-\frac{1}{2})\times2-3\times\frac{3}{4}}{2^{2}}=\frac{-1-\frac{9}{4}}{4}=\frac{-\frac{4 + 9}{4}}{4}=-\frac{13}{16}$.

Answer:

(a) $4$
(b) $-\frac{13}{16}$