Question

Let P(x)=F(x)G(x) and Q(x)=\frac{F(x)}{G(x)}, where F and G are the functions whose graphs are shown. a) Find P'(2) b) Find Q'(7)

Let \(P(x)=F(x)G(x)\) and \(Q(x)=\frac{F(x)}{G(x)}\), where F and G are the functions whose graphs are shown.
a) Find \(P'(2)\)
b) Find \(Q'(7)\)

Answers (1)

2021-06-04
Step 1
a) \(P'(2)\) will be the derivative of P(x) evaluated at \(x=2\). So first, we take a derivative.
In this section, you should have learned product rule, so \(P'(x)\) will look like this:
\(P'(x)=F(x)G'(x)+F'(x)G(x)\)
Now we let \(x=2\)
\(P'(2)=F(2)G'(2)+F'(2)G(2)\)
Read these values off the graph.
We see that \(F(2)=3\) (When its x-value is 2, the curve F has a y-value of 3)
We see that \(G(2)=2\).
\(F'(2)\) will be the slope of the curve F at 2. A tangent line at \(F(2)\) would be horizontal, so \(F'(2)=0\).
\(G(2)\) has a slope of \(\frac{1}{2}\), so \(G'(2)=\frac{1}{2}\)
Now we substitute these values into \(P'(2)\) to solve:
\(P'(2)=(3)\left(\frac{1}{2}\right)+(0)(2)=\left(\frac{3}{2}\right)+(0)=\frac{3}{2}\)
Step 2
We're playing by the same rules as above-differentiate the equation, read the values off the graph, substitute the values into the function, and solve.
Differentiate (Quotient Rule):
\(Q'(x)=\frac{G(x)F'(x)-F(x)G'(x)}{G(x)^{2}}\)
Read the values off the graph:
\(G(7)=1\)
\(G'(7)=-\frac{2}{3}\)
\(F(7)=5\)
\(F'(7)=\frac{1}{4}\)
Substitute:
\(Q'(7)=\frac{[1\cdot\left(\frac{1}{4}\right)-5\cdot\left(-\frac{2}{3}\right)]}{[1^{2}]}\)
\(Q'(7)=\left[\left(\frac{1}{4}\right)+\left(\frac{10}{3}\right)\right]\)
\(Q'(7)=\left(\frac{43}{12}\right)=3.5833\)
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