question 12\n6 pts\nfind all values of x (if any) where the tangent line to the graph of the function is…

question 12\n6 pts\nfind all values of x (if any) where the tangent line to the graph of the function is horizontal.\ny = 4x³ - 2x² + 2\nedit view insert format tools table\n12pt paragraph b i u a t²
Answer
Explanation:
Step1: Find the derivative
The derivative of $y = 4x^{3}-2x^{2}+2$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$ is $y'=12x^{2}-4x$.
Step2: Set the derivative equal to 0
A horizontal tangent line has a slope of 0. So we set $y' = 0$, i.e., $12x^{2}-4x=0$.
Step3: Factor out the common factor
Factor out $4x$ from the left - hand side: $4x(3x - 1)=0$.
Step4: Solve for x
Using the zero - product property, if $ab = 0$, then either $a = 0$ or $b = 0$. So $4x=0$ gives $x = 0$, and $3x-1=0$ gives $x=\frac{1}{3}$.
Answer:
$x = 0,x=\frac{1}{3}$