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

2 Answers
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]}

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)