## Calculus: Early Transcendentals (2nd Edition)

$= \frac{{ - 1 - {x^2}a - x{a^2}}}{{xa}}$
$\begin{gathered} f\,\left( x \right) = \frac{1}{x} - {x^2} \hfill \\ use\,\,the\,\,formula\,\,\frac{{f\,\left( x \right) - f\,\left( a \right)}}{{x - a}} \hfill \\ \hfill \\ \frac{{f\,\left( x \right) - f\,\left( a \right)}}{{x - a}} = \frac{{\frac{1}{x} - {x^2} - \frac{1}{a} + {a^2}}}{{x - a}} \hfill \\ \hfill \\ simplify\,\,\,the\,\,numerator \hfill \\ \hfill \\ = \frac{{\frac{{a - {x^3}a - x + {a^3}x}}{{xa}}}}{{x - a}} \hfill \\ \hfill \\ factor \hfill \\ \hfill \\ = \frac{{\frac{{a - x - xa\,\left( {{x^2} - {a^2}} \right)}}{{xa}}}}{{x - a}} \hfill \\ \hfill \\ = \frac{{\frac{{a - x - xa\,\left( {x - a} \right)\,\left( {x + a} \right)}}{{xa}}}}{{x - a}} \hfill \\ \hfill \\ = \frac{{\frac{{a - x + xa\,\left( {a - x} \right)\,\left( {x + a} \right)}}{{xa}}}}{{x - a}} \hfill \\ \hfill \\ = \frac{{\frac{{\,\left( {a - x} \right)\,\left( {1 + xa\,\left( {x + a} \right)} \right)}}{{xa}}}}{{x - a}} \hfill \\ \hfill \\ = \frac{{\,\left( {a - x} \right)\,\left( {1 + xa\,\left( {x + a} \right)} \right)}}{{xa\,\left( {x - a} \right)}} \hfill \\ \hfill \\ cancel\,\,\,x - a \hfill \\ \hfill \\ = \frac{{ - \,\left( {1 + xa\,\left( {x + a} \right)} \right)}}{{xa}} \hfill \\ \hfill \\ = \frac{{ - 1 - xa\,\left( {x + a} \right)}}{{xa}} \hfill \\ \hfill \\ = \frac{{ - 1 - {x^2}a - x{a^2}}}{{xa}} \hfill \\ \end{gathered}$