## Calculus: Early Transcendentals 8th Edition

(a) $$P'(2)=\frac{3}{2}$$ (b) $$Q'(7)=\frac{43}{12}$$
(a) $P(x)=F(x)G(x)$. Therefore, $$P'(x)=F'(x)G(x)+F(x)G'(x)$$ That means, $$P'(2)=F'(2)G(2)+F(2)G'(2)$$ From the graph, we can see that $F(2)=3$ and $G(2)=2$ To find $F'(2)$, we need to find the slope of the tangent line to $F$ at $x=2$ However, we see that the point where $x=2$ is the vertex (lowest point) of a small parabola in graph $F$. That means the tangent line there would be parallel with the $Ox$ line, and its slope would be $0$. In other words, $F'(2)=0$ Similarly, $G'(2)$ is also the slope of tangent line to $G$ at $x=2$. From $x=0$ to $x=4$, graph $G$ is a straight line. So, $$G'(2)=\frac{G(4)-G(0)}{4-0}=\frac{3-1}{4}=\frac{1}{2}$$ Therefore, $$P'(2)=0\times2+3\times(\frac{1}{2})=\frac{3}{2}$$ (b) $Q(x)=F(x)/G(x)$. Therefore, $$Q'(x)=\frac{F'(x)G(x)-F(x)G'(x)}{[G(x)]^2}$$ That means, $$Q'(7)=\frac{F'(7)G(7)-F(7)G'(7)}{[G(7)]^2}$$ From the graph, we can see that $F(7)=5$ and $G(7)=1$ To find $F'(7)$, we need to find the slope of the tangent line to $F$ at $x=7$ However, from $x=3$ to $x=7$, graph $F$ is a straight line. So, the slope of the tangent line to $F$ at $x=7$ is $$F'(7)=\frac{F(7)-F(3)}{7-3}=\frac{5-4}{4}=\frac{1}{4}$$ Similarly, $G'(7)$ is also the slope of tangent line to $G$ at $x=7$. From $x=4$ to $x=7$, graph $G$ is also a straight line. So, $$G'(7)=\frac{G(7)-G(4)}{7-4}=\frac{1-3}{3}=\frac{-2}{3}$$ Therefore, $$Q'(7)=\frac{1/4\times1-5\times(-2/3)}{1^2}=\frac{1/4+10/3}{1}=\frac{43}{12}$$