Question

find any critical numbers of the function. (enter your answers as a comma - separated list.)
g(x)=x^{6}-3x^{4}
x =

Explanation:

Step1: Differentiate the function

Using the power rule \((x^n)^\prime = nx^{n - 1}\), if \(g(x)=x^{6}-3x^{4}\), then \(g^\prime(x)=6x^{5}-12x^{3}\).

Step2: Set the derivative equal to zero

Set \(g^\prime(x) = 0\), so \(6x^{5}-12x^{3}=0\). Factor out \(6x^{3}\): \(6x^{3}(x^{2}-2)=0\).

Step3: Solve for \(x\)

From \(6x^{3}=0\), we get \(x = 0\). From \(x^{2}-2=0\), we have \(x^{2}=2\), then \(x=\pm\sqrt{2}\).

Answer:

\(0,-\sqrt{2},\sqrt{2}\)