f(x) = 3*x^5 - 5*x^3 - 8
f ' (x) = 15x^4 - 15x^2
= 0 for a max/min
15x^4 - 15x^2 = 0
x^4 - x^2 = 0
x^2(x^2 - 1) = 0
x = 0, 1, or -1
f(0) = -8
f(1) = 3 - 5 - 8 = -10
f(-1) = 3(-1) - 5(-1) - 8
= -6
I will let you decide how those 3 values of x affect the max/mins
Perhaps using something like
h tt ps://www.desmos.com/calculator
will help with that
btw, delete those two spaces at the front of the URL
For what x coordinates(s) does the function f(x) = 3*x^5 - 5*x^3 - 8 have a relative minimum?
2 answers
f"(x) = 60x^3 - 30x = 30x (2x^2 - 1)
when f'(x) = 0
f(x) has a max if f"(x) < 0 (concave downward)
f(x) has a min if f"(x) > 0 (concave upward)
when f'(x) = 0
f(x) has a max if f"(x) < 0 (concave downward)
f(x) has a min if f"(x) > 0 (concave upward)