Question

find the x - value(s) on the graph of f(x)=2x^3 + 6x^2 - 144x + 15 where the tangent is horizontal. 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 $f(x)=2x^{3}+6x^{2}-144x + 15$ using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ is $f'(x)=6x^{2}+12x-144$.

Step2: Set the derivative equal to 0

A horizontal tangent has a slope of 0. So we set $f'(x) = 0$, i.e., $6x^{2}+12x - 144=0$. Divide through by 6 to simplify: $x^{2}+2x - 24=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}+2x - 24=(x + 6)(x - 4)=0$. Then, using the zero - product property, if $(x + 6)(x - 4)=0$, then $x+6 = 0$ or $x - 4=0$.

Step4: Find the x - values

Solving $x+6 = 0$ gives $x=-6$, and solving $x - 4=0$ gives $x = 4$.

Answer:

$-6,4$