Question

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

Answer 1

`2y = x -3`

Explanation:

First find the gradient of `f(x)` by differentiating. Then if the gradient is `m` the slope of the normal is `-1/m`
The rest of the equation can be found by substituting in the values at the given point.

`f(x) = lnx - x^3`
`f'(x) = 1/x -3x^2`
At `x=1`, `m = 1 - 3.1^2 = -2`

Therefore the slope of the normal is `1/2`

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

The equation of the normal is `y=mx+c`
At the given point `-1 = 1/2(1) +c`
`:. c = -3/2`

The equation of the normal is
`y = 1/2x - 3/2`
or `2y = x -3`