Question

What is the equation of the normal line of `f(x)=e^x-x^3` at `x=0`?

Answer 1

In slope intercept form:

`y = -x+1`

Explanation:

Given `f(x) = e^x-x^3`

Then `f'(x) = e^x-3x^2`

`f(0) = e^0 - 0 = 1`

So the graph of the function passes through `(0, 1)`

`f'(0) = e^0 - 0 = 1`

So the slope of the tangent at `(0, 1)` is `1`.

If the slope of the tangent is `m` then the slope of the normal is `-1/m`. So in our example the slope of the normal is `-1/1 = -1`.

The equation of a line of slope `m` passing through a point `(x_0, y_0)` can be written as:

`(y - y_0) = m(x - x_0)`

In our case we have `(x_0, y_0) = (0, 1)` and `m = -1`, so:

`(y - 1) = -1(x - 0) = -x`

That is `y = -x+1` in slope intercept form.

graph{(e^x-x^3-y)(x+y-1) = 0 [-10, 10, -5, 5]}