## Calculus: Early Transcendentals 8th Edition

(a) $$u'(1)=0$$ (b) $$v'(5)=\frac{-2}{3}$$
(a) $u(x)=f(x)g(x)$. Therefore, $$u'(x)=f'(x)g(x)+f(x)g'(x)$$ That means, $$u'(1)=f'(1)g(1)+f(1)g'(1)$$ From the graph, we can easily see that $f(1)=2$ and $g(1)=1$ To find $f'(1)$, we need to find the slope of the tangent line to $f$ at $x=1$ However, from $x=0$ to $x=2$, graph $f$ is a straight line. So, the slope of the tangent line to $f$ at $x=1$ is $$f'(1)=\frac{f(2)-f(0)}{2-0}=\frac{4-0}{2-0}=2$$ Similarly, $g'(1)$ is also the slope of tangent line to $g$ at $x=1$. Also, from $x=0$ to $x=2$, graph $g$ is also a straight line. So, $$g'(1)=\frac{g(2)-g(0)}{2-0}=\frac{0-2}{2-0}=-1$$ Therefore, $$u'(1)=2\times1+2\times(-1)=0$$ (b) $v(x)=f(x)/g(x)$. Therefore, $$v'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$ That means, $$v'(5)=\frac{f'(5)g(5)-f(5)g'(5)}{[g(5)]^2}$$ From the graph, we can easily see that $f(5)=3$ and $g(5)=2$ To find $f'(5)$, we need to find the slope of the tangent line to $f$ at $x=5$ However, from $x=2$ to $x=5$, graph $f$ is a straight line. So, the slope of the tangent line to $f$ at $x=5$ is $$f'(5)=\frac{f(5)-f(2)}{5-2}=\frac{3-4}{3}=\frac{-1}{3}$$ Similarly, $g'(5)$ is also the slope of tangent line to $g$ at $x=5$. Also, from $x=2$ to $x=5$, graph $g$ is also a straight line. So, $$g'(5)=\frac{g(5)-g(2)}{5-2}=\frac{2-0}{3}=\frac{2}{3}$$ Therefore, $$v'(5)=\frac{-1/3\times2-3\times(2/3)}{2^2}=\frac{-2/3-2}{4}=\frac{-2}{3}$$