SOLUTION: can someone help me solve this?? ƒ(x) = (x4)/4 − 3x3 + (23x2)/2 - 15x Find the x values where the extremes occur. A. 1, -3, 5 B. 1, 3, -5 C. 1, 3, 5 D. 1, -3

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: can someone help me solve this?? ƒ(x) = (x4)/4 − 3x3 + (23x2)/2 - 15x Find the x values where the extremes occur. A. 1, -3, 5 B. 1, 3, -5 C. 1, 3, 5 D. 1, -3      Log On


   

Question 1028960: can someone help me solve this??
ƒ(x) = (x4)/4 − 3x3 + (23x2)/2 - 15x
Find the x values where the extremes occur.

A. 1, -3, 5
B. 1, 3, -5
C. 1, 3, 5
D. 1, -3, -5
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
using calculus and taking the derivative
x^3-9x^2+23x-15=0
Test each.
1 works, and it does for all choices
for 5, 125-225+115-15=0, so 5 works.
for 3, 27-81+69-15=0
It is C.
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