# What is the slope of the line normal to the tangent line of f(x) = x^3-4x  at  x= -1 ?

Feb 28, 2016

$1$

#### Explanation:

There is a multi-step process that goes into finding the slope of the normal line:

• Find the function's derivative.
• Find the value of the derivative at the desired point. This is a slope of the tangent line.
• Take the opposite reciprocal of the slope of the tangent line. This is the slope of the normal line since the normal line and tangent line are perpendicular.

Find the derivative through the power rule:

$f \left(x\right) = {x}^{3} - 4 x$

$f ' \left(x\right) = 3 {x}^{2} - 4$

The slope of the tangent line is

$f ' \left(- 1\right) = 3 {\left(- 1\right)}^{2} - 4 = 3 \left(1\right) - 4 = \textcolor{b l u e}{- 1}$

The slope of the normal line is

$- \left(\frac{1}{\textcolor{b l u e}{- 1}}\right) = 1$