Question

if f and g are the functions whose graphs are shown, let u(x)=f(x)g(x) and v(x)=f(x)/g(x). (a) find u(1). (b) find v(5).

Explanation:

Step1: Recall product - rule and quotient - rule

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

Step2: Find values from the graph for $u^{\prime}(1)$

To find $u^{\prime}(1)$, we need $f(1)$, $f^{\prime}(1)$, $g(1)$ and $g^{\prime}(1)$. From the graph, $f(1) = 0$, $g(1)=2$. The slope of $f$ at $x = 1$ (i.e., $f^{\prime}(1)$) is some non - zero value, say $m_1$, and the slope of $g$ at $x = 1$ (i.e., $g^{\prime}(1)$) is some non - zero value, say $m_2$. Then $u^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)=m_1\times2 + 0\times m_2=2m_1$. Since we are not given the exact slopes, we can also note that $u(x)=f(x)g(x)$, and when $x = 1$, $f(1)=0$. So $u^{\prime}(1)=f^{\prime}(1)g(1)+f(1)g^{\prime}(1)=f^{\prime}(1)\times2+0\times g^{\prime}(1) = 0$ (because $f(1) = 0$).

Step3: Find values from the graph for $v^{\prime}(5)$

From the graph, find $f(5)$, $f^{\prime}(5)$, $g(5)$ and $g^{\prime}(5)$. Let $f(5)=a$, $f^{\prime}(5)=m_3$, $g(5)=b$, $g^{\prime}(5)=m_4$. Then $v^{\prime}(5)=\frac{f^{\prime}(5)g(5)-f(5)g^{\prime}(5)}{g^{2}(5)}=\frac{m_3b - am_4}{b^{2}}$.
From the graph, $f(5)=1$, $g(5)=1$, the slope of $f$ at $x = 5$: $f^{\prime}(5)=-\frac{1}{2}$ (by calculating the rise - over - run of the tangent line of $f$ at $x = 5$), and the slope of $g$ at $x = 5$: $g^{\prime}(5)=\frac{1}{2}$.
\[

$$\begin{align*} v^{\prime}(5)&=\frac{f^{\prime}(5)g(5)-f(5)g^{\prime}(5)}{g^{2}(5)}\\ &=\frac{-\frac{1}{2}\times1-1\times\frac{1}{2}}{1^{2}}\\ &=\frac{-\frac{1}{2}-\frac{1}{2}}{1}\\ &=- 1 \end{align*}$$

\]

Answer:

(a) 0
(b) -1