Question

for the function, find the points on the graph at which the tangent line is horizontal. if none exist, state that fact.
f(x)=4x^{2}-2x + 4
select the correct choice below and, if necessary, fill in the answer box within your choice.
a. the point(s) at which the tangent line is horizontal is (are)
(simplify your answer. type an ordered - pair. use a comma to separate answers as needed.)
b. there are no points on the graph where the tangent line is horizontal
c. the tangent line is horizontal at all points of the graph

Explanation:

Step1: Find the derivative

The derivative of $f(x)=4x^{2}-2x + 4$ using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=8x-2$.

Step2: Set the derivative equal to 0

A horizontal tangent line has a slope of 0. So we set $f'(x)=0$.
$8x - 2=0$.

Step3: Solve for x

Add 2 to both sides: $8x=2$. Then divide both sides by 8, we get $x=\frac{2}{8}=\frac{1}{4}$.

Step4: Find the y - value

Substitute $x = \frac{1}{4}$ into the original function $f(x)=4x^{2}-2x + 4$.
$f(\frac{1}{4})=4(\frac{1}{4})^{2}-2(\frac{1}{4})+4=4\times\frac{1}{16}-\frac{2}{4}+4=\frac{1}{4}-\frac{1}{2}+4=\frac{1 - 2+16}{4}=\frac{15}{4}$.

Answer:

A. $(\frac{1}{4},\frac{15}{4})$