Question

Use equation 5 to find \(f'\left( a \right)\) at the given number \(a\).

21. \(f\left( x \right) = \frac{{{x^2}}}{{x + 6}}\), \(a = {\bf{3}}\)

Short answer

The value of \(f'\left( a \right)\) at \(a = 3\) is \(\frac{5}{9}\).

Step-by-step solution

Write equation 5

The equation is shown below:

\(f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\)

The value of the function

Substitute \(a = 3\) in equation 5 as follows:

\(\begin{aligned}f'\left( 3 \right)& = \mathop {\lim }\limits_{x \to 3} \frac{{f\left( x \right) - f\left( 3 \right)}}{{x - 3}}\\ &= \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{{x^2}}}{{x + 6}} - \frac{{{{\left( 3 \right)}^2}}}{{3 + 6}}}}{{x - 3}}\\ &= \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{{x^2}}}{{x + 6}} - \frac{9}{9}}}{{x - 3}}\\ &= \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{{x^2}}}{{x + 6}} - 1}}{{x - 3}}\\& = \mathop {\lim }\limits_{x \to 3} \frac{{\frac{{{x^2} - \left( {x + 6} \right)}}{{x + 6}}}}{{x - 3}}\end{aligned}\)

Solve the above equation further,

\(\begin{aligned}f'\left( 3 \right) &= \mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - x - 6}}{{\left( {x + 6} \right)\left( {x - 3} \right)}}\\ &= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x + 2} \right)\left( {x - 3} \right)}}{{\left( {x + 6} \right)\left( {x - 3} \right)}}\\ &= \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x + 2} \right)}}{{\left( {x + 6} \right)}}\\ & = \frac{{3 + 2}}{{3 + 6}}\\ &= \frac{5}{9}\end{aligned}\)

The value of \(f'\left( a \right)\) at \(a = 3\) is \(\frac{5}{9}\).