Question

for what value(s) of x does the graph of f(x)=6x^3 - 9x^2 - 216x have a horizontal tangent? if there is more than one x - value, enter them in a comma - separated list. if an x - value does not exist, enter dne.
x =

Explanation:

Step1: Find the derivative

The derivative of $f(x)=6x^{3}-9x^{2}-216x$ using the power - rule $(x^n)' = nx^{n - 1}$ is $f'(x)=18x^{2}-18x - 216$.

Step2: Set the derivative equal to zero

A horizontal tangent occurs when $f'(x) = 0$. So we set $18x^{2}-18x - 216=0$. Divide through by 18 to simplify: $x^{2}-x - 12 = 0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}-x - 12=(x - 4)(x+3)=0$. Then, using the zero - product property, if $(x - 4)(x + 3)=0$, then $x-4 = 0$ or $x + 3=0$.

Answer:

$x=-3,4$