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:

\(Q'(7)=\frac{[1\cdot\left(\frac{1}{4}\right)-5\cdot\left(-\frac{2}{3}\right)]}{[1^{2}]}\)

\(Q'(7)=\left[\left(\frac{1}{4}\right)+\left(\frac{10}{3}\right)\right]\)

\(Q'(7)=\left(\frac{43}{12}\right)=3.5833\)

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:

\(Q'(7)=\frac{[1\cdot\left(\frac{1}{4}\right)-5\cdot\left(-\frac{2}{3}\right)]}{[1^{2}]}\)

\(Q'(7)=\left[\left(\frac{1}{4}\right)+\left(\frac{10}{3}\right)\right]\)

\(Q'(7)=\left(\frac{43}{12}\right)=3.5833\)