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)?

Mar 9, 2017

Relative minimum at $x = - 1$.

Explanation:

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

$f \left(x\right) = - 2 \left(x + 2\right) \left(x - 1\right) \left(x - 1\right)$

$f \left(x\right) = - 2 \left(x + 2\right) \left({x}^{2} - 2 x + 1\right)$

$f \left(x\right) = - 2 \left({x}^{3} + 2 {x}^{2} - 2 {x}^{2} - 4 x + x + 2\right)$

$f \left(x\right) = - 2 \left({x}^{3} - 3 x + 2\right)$

$f \left(x\right) = - 2 {x}^{3} + 6 x - 4$

We now differentiate using the power rule.

$f ' \left(x\right) = - 6 {x}^{2} + 6$

$f ' \left(x\right) = 6 \left(1 - {x}^{2}\right)$

$f ' \left(x\right) = 6 \left(1 + x\right) \left(1 - x\right)$

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.

$0 = 6 \left(1 + x\right) \left(1 - x\right)$

$x = - 1 \mathmr{and} 1$

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 $\left(- 3 , 3\right)$

Test point: $x = - 2$

$f \left(- 2\right) = - 6 {\left(- 2\right)}^{2} + 6 = - 6 \left(4\right) + 6 = - 18$

Test point: $x = 0$

$f \left(0\right) = - 6 {\left(0\right)}^{2} + 6 = 6$

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

Hopefully this helps!