# What is the equation of the line normal to  f(x)=lnx-x^3 at  x=1?

Jan 15, 2016

$2 y = x - 3$

#### Explanation:

First find the gradient of $f \left(x\right)$ by differentiating. Then if the gradient is $m$ the slope of the normal is $- \frac{1}{m}$
The rest of the equation can be found by substituting in the values at the given point.

$f \left(x\right) = \ln x - {x}^{3}$
$f ' \left(x\right) = \frac{1}{x} - 3 {x}^{2}$
At $x = 1$, $m = 1 - {3.1}^{2} = - 2$

Therefore the slope of the normal is $\frac{1}{2}$

Going back to the original expression
f(1) = ln(1) - 1^3 = 0 -1 = -1#

The equation of the normal is $y = m x + c$
At the given point $- 1 = \frac{1}{2} \left(1\right) + c$
$\therefore c = - \frac{3}{2}$

The equation of the normal is
$y = \frac{1}{2} x - \frac{3}{2}$
or $2 y = x - 3$