Question

determine all critical points for the function.
f(x)=80x^{3}-3x^{5}

a. x = - 4 and x = 4
b. x = 0, x = - 4, and x = 4
c. x = - 4
d. x = 4

Explanation:

Step1: Find the derivative of the function

Use the power rule \((x^n)^\prime = nx^{n - 1}\).
For \(y = 80x^{3}-3x^{5}\), the derivative \(y^\prime=f^\prime(x)=80\times3x^{2}-3\times5x^{4}=240x^{2}-15x^{4}\).

Step2: Set the derivative equal to zero

Set \(240x^{2}-15x^{4}=0\).
Factor out \(15x^{2}\): \(15x^{2}(16 - x^{2})=0\).
Using the zero - product property \(ab = 0\) implies \(a = 0\) or \(b = 0\).
First, \(15x^{2}=0\) gives \(x = 0\).
Second, \(16 - x^{2}=0\), which can be written as \(x^{2}=16\), so \(x=\pm4\).

Answer:

B. \(x = 0,x=-4,\text{ and }x = 4\)