Find the relative extrema?

Question

of 3x^5 - 5X^3

Answer 1

We have:
f'(x) = 15x^4 - 15x^2 = 15x^2(x+1)(x-1)
f''(x) = 60x^3 - 30x = 30x (2x^2 - 1)

Solving f'(x) = 0 yields the stationary points x = 0, x = -1, and x=1. As shown in the following table, we can conclude from the second derivative test that f has a relative maximum at x = -1 and a relative minimum at x = 1.

Stationary Point - 30x(2x^2 -1) - f''(x) 2ndDerivTest
x = -1 -30 - f has rel max
x = 0 0 0 Inconclusive
x = 1 30 + f has rel min

The test is inconclusive at x = 0, so we will try thte first derivative test at that point. A sign analysis of f' is given in the following table:

Interval - 15x^2(x+1)(x-1) - f'(x)
-1 0
Since there is no sign change in f' at x = 0, there is neither a relative maximum nor a relative minimum at that point. All of this is consistent with the graph, if you use a graphing calculator...

hope that helps!!!

Answer 2

Relative extrema (maximum and minimum points) occur when the derivative equals zero. The derivative of this function is
15x^4 - 15x^2
If you set that equal to 0 and factor, you have
15x^2(x^2 - 1) = 0
15x^2(x - 1)(x + 1) = 0
The relative extrama would occur at x=0 and x=1 x=-1.
Technically extrema are points, so you need to find the corresponding y-values by plugging into the function you are given.

Answer 3

y=3x^5-5x^3
y'=15x^4-15x^2
settingit to zero15x^2(x^2-1)=0
either x=0 or x^2-1=0
which means (x+1)(x-1)=0
or x=+/-1
y''=60x^3-30x
at x=-1 y''=negative so local max
at x=1 y'=positive so local min