How do i find the values of x that give relative extrema for the function f(x)=3x^5-5x^3?

Question

i need to learn how to do this, so if u can solve it please show work.

Answer 1

f(x) = 3x^5 - 5x^3
f'(x) = 15x^4 - 15x^2 = 0
x^2(x^2 - 1) = 0
x^2(x + 1)(x - 1) = 0
x = -1,0,1
f''(x) = 60x^3 - 30x
inflection points are:
x(2x^2 - 1) = 0
x = 0, ±√(1/2)
so, there is a relative maximum @ x = -1 and a relative minimum @ x = 1. x = 0 is simply a point at which the slope is 0 and changes sign.

checking,
f(-2) = -96 + 40 = -56
f(-1) = -3 + 5 = 2
f(-1/2) = -3/32 + 5/8 = 17/32
f(0) = 0
f(1/2) = 3/32 - 5/8 = -17/32
f(1) = 3 - 5 = -2
f(2) = 96 - 40 = 56

Answer 2

you should crate F' (x).

F(x) = 3x^5 - 5x^3

F'(x) = 15x^4 - 15x^2
F'(x) = 15x^2(x - 1)

after that you should find x that F'(x) = 0


F'(x) = 0

15x^2(x^2 - 1) = 0

it men:

1. 15x^2 = 0 => x^2=0 => X = 0

2. x^2 - 1 = 0 => x^2 = 1 => x = 1 and -1

so for x =(( 0 & +1 & -1)) , F'(x) = 0

just now you should find out which one is your answer. this part is your job. just you should know that the answer is



X = -1