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
Mar 9, 2017

Answer:

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!