determine all critical points for the function.\nf(x)=80x^{3}-3x^{5}\n\na. x = - 4 and x = 4\nb. x = 0, x =…

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)