Question

Given `f(x) = -2(x+2)(x-1)^2` on the open interval (-3,3). How do you determine the x coordinate of the relative minimum of f (x) in the open interval (-3,3)?

Answer 1

Relative minimum at `x = -1`.

Explanation:

Start by differentiating. Since we're not dealing with too high powers, it would be easiest to expand.

`f(x) = -2(x + 2)(x - 1)(x - 1)`

`f(x) = -2(x + 2)(x^2 - 2x + 1)`

`f(x) = -2(x^3 + 2x^2 - 2x^2 - 4x + x + 2)`

`f(x) = -2(x^3 - 3x + 2)`

`f(x) = -2x^3 + 6x - 4`

We now differentiate using the power rule.

`f'(x) = -6x^2 + 6`

`f'(x) = 6(1 - x^2)`

`f'(x) = 6(1 + x)(1 - x)`

We now find the critical points. These occur whenever the derivative is `0` or undefined. However, this is a polynomial function and is defined on all values of `x`. If we set the derivative to `0`, we have.

`0 = 6(1 + x)(1 - x)`

`x = -1 and 1`

We must now test the sign of the derivative on both sides of these points.

`•f'(a) < 0` at a point `x = a`, then the function is decreasing at `x = a`
`•f'(a) > 0` at a point `x = a`, then the function is increasing at `x = a`.

If the function is decreasing on one side of a critical point and increasing on the other side, then we have an absolute minimum on `(-3, 3)`

Test point: `x = -2`

`f(-2) = -6(-2)^2 + 6 = -6(4) + 6 = -18`

Test point: `x = 0`

`f(0) = -6(0)^2 + 6 = 6`

Therefore, our relative minimum will be at `x = -1`. There will be no absolute minimum in this open interval because the function approaches `-oo` to the right and `+oo` on the left.

Hopefully this helps!