Question

Let \(P\left( x \right) = F\left( x \right)G\left( x \right)\) and \(Q\left( x \right) = \frac{{F\left( x \right)}}{{G\left( x \right)}}\), where \(F\) and \(G\) are

the functions whose graphs are shown.

(a) Find \(P'\left( 2 \right)\) (b) Find \(Q'\left( 7 \right)\)

Short answer

(a) The required value is \(P'\left( 2 \right) = \frac{3}{2}\).

(b) The required value is \(Q'\left( 7 \right) = \frac{{43}}{{12}}\).

Step-by-step solution

Determine the slope of a line segment passing through a point

The slope of a line segment passes through the point \(\left( {c,f\left( c \right)} \right)\) with endpoints \(\left( {a,f\left( a \right)} \right)\) and \(\left( {b,f\left( b \right)} \right)\) is defined by \(f'\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}}\).

Finding the derivative of a function at a point

(a) Given that \(P\left( x \right) = F\left( x \right)G\left( x \right)\).

Then we get by differentiating,

\(P'\left( x \right) = F\left( x \right)G'\left( x \right) + G\left( x \right)F'\left( x \right)\).

Now to find \(P'\left( 2 \right)\) we have to find \(F'\left( 2 \right)\) and \(G'\left( 2 \right)\) first.

The derivative of \(F\) at the point \(x = 2\) is \(F'\left( 2 \right) = 0\).

Also, since the line segment with the points \(\left( {2,G\left( 2 \right)} \right)\) and \(\left( {4,G\left( 4 \right)} \right)\) passing through the point \(\left( {2,G\left( 2 \right)} \right)\) so the derivative at the point \(x = 2\) will be,

\(\begin{aligned}G'\left( 2 \right) & = \frac{{G\left( 4 \right) - G\left( 2 \right)}}{2}\\ & = \frac{{3 - 2}}{2}\\ & = \frac{1}{2}\end{aligned}\)

Now since \(F\left( 2 \right) = 3\) and \(G\left( 2 \right) = 2\) so derivative of the function \(P\) at the point \(x = 2\) will be,

\(\begin{aligned}P'\left( 2 \right) & = F\left( 2 \right)G'\left( 2 \right) + G\left( 2 \right)F'\left( 2 \right)\\ & = 3 \times \frac{1}{2} + 2 \times 0\\ & = \frac{3}{2}\end{aligned}\)

Hence \(P'\left( 2 \right) = \frac{3}{2}\).

Finding the derivative of a function at a point

Given that \(Q\left( x \right) = \frac{{F\left( x \right)}}{{G\left( x \right)}}\).

Then we get by differentiating,

\(Q'\left( x \right) = \frac{{G\left( x \right)F'\left( x \right) - F\left( x \right)G'\left( x \right)}}{{{{\left( {G\left( x \right)} \right)}^2}}}\).

Now to find \(Q'\left( 7 \right)\) we have to find \(F'\left( 7 \right)\) and \(G'\left( 7 \right)\) first.

Since the line segment with the points \(\left( {3,F\left( 3 \right)} \right)\) and \(\left( {7,F\left( 7 \right)} \right)\) passing through the point \(\left( {7,F\left( 7 \right)} \right)\) so the derivative of \(F\) at the point \(x = 7\) will be,

\(\begin{aligned}F'\left( 7 \right) & = \frac{{F\left( 7 \right) - f\left( 3 \right)}}{4}\\ & = \frac{{5 - 4}}{4}\\ & = \frac{1}{4}\end{aligned}\)

Since for the graph \(G\) line segment with \(\left( {4,G\left( 4 \right)} \right)\) and \(\left( {7,G\left( 7 \right)} \right)\) passing through the point \(\left( {7,G\left( 7 \right)} \right)\), so the slope at \(x = 7\) will be,

\(\begin{aligned}G'\left( 7 \right) & = \frac{{G\left( 7 \right) - G\left( 4 \right)}}{3}\\ & = \frac{{1 - 3}}{3}\\ & = - \frac{2}{3}\end{aligned}\)

Now since \(F\left( 7 \right) = 5\) and \(G\left( 7 \right) = 1\) so derivative of the function \(Q\) at the point \(x = 7\) will be,

\(\begin{aligned}Q'\left( 7 \right) & = \frac{{G\left( 7 \right)F'\left( 7 \right) - F\left( 7 \right)G'\left( 7 \right)}}{{{{\left( {G\left( 7 \right)} \right)}^2}}}\\ & = \frac{{1 \times \frac{1}{4} - 5 \times \left( { - \frac{2}{3}} \right)}}{1}\\ & = \frac{1}{4} + \frac{{10}}{3}\\ & = \frac{{43}}{{12}}\end{aligned}\)

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