Question

For `f(x) =(x^2 + 3x - 10)^3`, what is the equation of the line tangent to `x =5` and at what point is the tangent line horizontal?

Answer 1

The tangent line will be horizontal at `x = -3/2, -5 and 2`.

Explanation:

This is essentially two problems in one. I will answer where the tangent line is horizontal.

We start by differentiating `f(x)`, using the chain rule.

Letting `y= u^3` and `u = x^2 + 3x - 10`. Then dy/(du) = 3u^2`and`(du)/dx = 2x + 3#

Then:

`dy/dx = dy/(du) xx (du)/dx`

`dy/dx = 3u^2 xx 2x + 3`

`dy/dx = (6x + 9)(x^2 + 3x - 10)^2`

The tangent will be horizontal if its slope is `0`. The derivative represents the rate of change of the function at any given point in its domain.

We set the derivative to `0` and solve.

`0 = (6x + 9)(x^2 + 3x - 10)^2`

`0 = 6x + 9 and 0 = x^2 + 3x -10`

`x = -9/6, 2, -5 = -3/2, 2, -5`

Hopefully this helps!