What are the relative extrema of this equation? x^4 - 2x^3

2 Answers
Sam F.
Feb 26, 2018

Local minimum at #(3/2, -27/16)#

Explanation:

To find local extrema, we use the first an second derivative tests.

#f(x)=x^4-2x^3#
#f'(x)=4x^3-6x^2#

For the sake of the first derivative test, let's factor this to:

#f'(x)=2x^2(2x-3)#

If we set #f'(x)=0#, we find two possible extrema at #x=0# and #x=3/2#

Now we use the second derivative test to determine minimum/maximum/point of inflection:

#f'(x)=4x^3-6x^2#
#f''(x)=12x^2-12x#

Factoring, we get

#f''(x)=12x(x-1)#

Now we evaluate our two possible extrema using the second derivative test:

#f''(0)=12(0)(0-1)=0# This is a point of inflection, not an extreme.

#f''(3/2)=12(3/2)(3/2-1)=9# The graph is concave up at #x=3/2# and is therefore a local minimum. Plugging this #x# back into our original function gives a local minimum at point #(3/2, -27/16)#

All of this can be observed on the graph of the original function:

graph{x^4-2x^3 [-1.49, 4.67, -1.958, 1.122]}

VNVDVI
Feb 26, 2018

We have a local minimum at #(3/2, -27/16)#

Explanation:

Determine #f'(x)#, set #f'(x)=0#, solve for #x# and determine the values of #x# for which #f'(x)# doesn't exist:

#f'(x)=4x^3-(3)(2)x^2=4x^3-6x^2#

#4x^3-6x^2=0#

#2x^2(2x-3)=0#

#2x^2=0:#

#x^2=0#
#x=0#

#2x-3=0:#

#2x=3#
#x=3/2#

#f'(x)# is a polynomial; therefore, it exists for all real #x#.

We have prospective extrema at #x=0, x=3/2#. Let's split up the domain of #f(x)#, which is #(-∞,∞)#, across these #x#-values. This gives us the intervals:

#(-∞,0),(0,3/2),(3/2,∞)#

At each of these intervals, we want to determine if #f'(x)# is positive or negative. We'll do this by selecting a value for #x# from each interval and plugging it into #f'(x)#. If #f'(x)>0# on an interval, #f(x)# is increasing on that interval. If #f'(x)<0# on an interval, #f(x)# is decreasing on that interval.

We have an extremum at any #x# value where #f'(x)# changes signs, IE, #f(x)# changes from increasing to decreasing (or vice versa).

#(-∞,0):#

Let's select #x=-1#.

#f'(-1)=4(-1)^3-6(-1^2)=-4-6<0#

#f(x)# is decreasing on #(-∞,0)#

#(0,3/2):#

Let's select #x=1.#

#f'(1)=4-6<0#

#f(x)# is also decreasing on #(0,3/2).#

#(3/2, ∞):#

Let's select #x=2.#

#f'(2)=4(2^3)-6(2^2)=4(8)-6(4)=32-24>0#

#f(x)# is increasing on #(3/2, ∞)#. This is a switch from the decrease on the previous two intervals, so we have an extremum at #x=3/2#. Since #f(x)# went from decreasing on #(0,3/2)# to increasing on #(3/2,∞),# we have a local minimum at #x=3/2.#

To determine the #y#-coordinate of our local minimum, find #f(3/2):#

#f(3/2)=(3/2)^4-2(3/8)^3=81/16-2(27/8)=81/16-54/8=81/16-108/16=-27/16#

We have a local minimum at #(3/2, -27/16)#

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