What is the slope of the line normal to the tangent line of f(x) = 1/x+x^3  at  x= 2 ?

Sep 17, 2016

$- \frac{4}{47}$

Explanation:

The slope of the line tangent to the curve $y = f \left(x\right)$ at $x = 2$ is given by $f ' \left(2\right)$.

First, find $f ' \left(x\right)$.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{x} + {x}^{3}\right)$

$= - \frac{1}{x} ^ 2 + 3 {x}^{2}$

Find $f ' \left(2\right)$ by substituting $x = 2$.

$f ' \left(2\right) = - \frac{1}{2} ^ 2 + 3 {\left(2\right)}^{2}$

$= \frac{47}{4}$

The slope of the line normal to the tangent line, $m$, is given by

$m = - \frac{1}{f ' \left(2\right)}$

$= - \frac{4}{47}$