Question

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

Answer 1

the gradient of the normal at `x=1` is `m=-1/5`

Explanation:

To answer this question we need to find the slope of the tangent when `x=1`; and use the fact that the normal and tangent are perpendicular the so the product of their slopes = -1

In order to this we need to evaluate `f'(1)` and hence we need to find `f'(x)`

`f(x)=e^(x^2-1)+3x-2`
`:. f'(x)=e^(x^2-1)(2x)+3`

We don't need to simplify any more - just substitute `x=1`:
`x=1 => f'(1)=e^(1-1)(2)(1)+3 = 5`

So the gradient of the tangent at `x=1` is `m=5`
So the gradient of the normal at `x=1` is `m=-1/5`