# 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)?

##### 1 Answer

#### Answer:

Relative minimum at

#### 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 = 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

**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

Hopefully this helps!