Question

If \(g\left( x \right) = xf\left( x \right)\), where \(f\left( 3 \right) = 4\)and \(f'\left( 3 \right) = - 2\), find an equation of the tangent line to the graph of \(g\) at the point where \(x = 3\)

Short answer

The equation of the tangent line is \(y = - 2x + 18\).

Step-by-step solution

Finding derivative when two functions are multiplied

Let the function be \(h\left( x \right) = f\left( x \right)g\left( x \right)\) then the derivative will be,

\(\begin{aligned}h'\left( x \right) & = \frac{d}{{dx}}f\left( x \right)g\left( x \right)\\ & = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + g\left( x \right)\frac{d}{{dx}}f\left( x \right)\end{aligned}\)

Also, the equation of the tangent line at point \(\left( {a,f\left( a \right)} \right)\) is \(y - f\left( a \right) = f'\left( a \right)\left( {x - a} \right)\).

Finding the derivative of the function at a given point

Given that \(g\left( x \right) = xf\left( x \right)\).

Differentiating the function we get

\(\begin{aligned}g'\left( x \right) & = \frac{d}{{dx}}xf\left( x \right)\\ & = xf'\left( x \right) + f\left( x \right)\end{aligned}\)

So \(g'\left( 3 \right) = 3 \times f'\left( 3 \right) + f\left( 3 \right)\).

Substitute 3 for x into the function of \(g'\left( x \right)\) to determine \(g'\left( 3 \right)\).

\(\begin{aligned}g'\left( 3 \right) & = 3 \times \left( { - 2} \right) + 4\\ & = - 6 + 4\\ & = - 2\end{aligned}\)

Substitute 3 for x into the function of \(g\left( x \right)\) to determine \(g\left( 3 \right)\).

\(\begin{aligned}g\left( 3 \right) & = 3 \times f\left( 3 \right)\\ & = 3 \times 4\\ & = 12\end{aligned}\)

Finding the equation of tangent line

Use the definition, the equation of the tangent line at point \(\left( {a,f\left( a \right)} \right)\) is \(y - f\left( a \right) = f'\left( a \right)\left( {x - a} \right)\)to find the equation of tangent line at the point \(x = 3\).

\(\begin{aligned}y - g\left( 3 \right) & = g'\left( 3 \right)\left( {x - 3} \right)\\y - 12 & = - 2\left( {x - 3} \right)\\y - 12 & = - 2x + 6\\y & = - 2x + 18\end{aligned}\)

Hence the equation of the tangent line is \(y = - 2x + 18\).