find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer…

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)\n\n$f(x)=2x^{3}-3x^{2}-36x$\n\n$x =$

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)\n\n$f(x)=2x^{3}-3x^{2}-36x$\n\n$x =$

Answer

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)