Question

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)

$f(x)=2x^{3}-3x^{2}-36x$

$x =$

Explanation:

Step1: Find the derivative of the function

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(y = f(x)=2x^{3}-3x^{2}-36x\), we have \(f^\prime(x)=2\times3x^{2}-3\times2x - 36\).
So, \(f^\prime(x)=6x^{2}-6x - 36\).

Step2: Set the derivative equal to zero

Set \(f^\prime(x)=0\), then \(6x^{2}-6x - 36 = 0\). Divide both sides of the equation by \(6\): \(x^{2}-x - 6=0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b=-1\), \(c = - 6\)), we use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or factor.
Factor \(x^{2}-x - 6=(x - 3)(x + 2)=0\).

Answer:

\(-2,3\)