Question

let ( f(x)=2x^{3}-0x^{2}-18x + 8)
a. find all points on the graph of ( f(x)) at which the tangent line is horizontal
b. find all points on the graph of ( f(x)) at which the tangent line has slope 192
a. the tangent line is horizontal at the point(s)
(type an ordered - pair. use a comma to separate answers as needed.)

Explanation:

Step1: Find the derivative of f(x)

The derivative of $f(x)=2x^{3}-0x^{2}-18x + 8$ (which is just $f(x)=2x^{3}-18x + 8$) using the power - rule $(x^n)'=nx^{n - 1}$ is $f'(x)=6x^{2}-18$.

Step2: Solve for when the tangent line is horizontal

A horizontal tangent line has a slope of 0. So we set $f'(x)=0$.
$6x^{2}-18 = 0$
Add 18 to both sides: $6x^{2}=18$
Divide both sides by 6: $x^{2}=3$
Take the square root of both sides: $x=\pm\sqrt{3}$
When $x = \sqrt{3}$, $f(\sqrt{3})=2(\sqrt{3})^{3}-18\sqrt{3}+8=2\times3\sqrt{3}-18\sqrt{3}+8=6\sqrt{3}-18\sqrt{3}+8=8 - 12\sqrt{3}$
When $x=-\sqrt{3}$, $f(-\sqrt{3})=2(-\sqrt{3})^{3}-18(-\sqrt{3})+8=-6\sqrt{3}+18\sqrt{3}+8=8 + 12\sqrt{3}$
The points are $(\sqrt{3},8 - 12\sqrt{3}),(-\sqrt{3},8 + 12\sqrt{3})$

Step3: Solve for when the slope of the tangent line is 192

Set $f'(x)=192$.
$6x^{2}-18 = 192$
Add 18 to both sides: $6x^{2}=210$
Divide both sides by 6: $x^{2}=35$
Take the square root of both sides: $x=\pm\sqrt{35}$
When $x=\sqrt{35}$, $f(\sqrt{35})=2(\sqrt{35})^{3}-18\sqrt{35}+8=2\times35\sqrt{35}-18\sqrt{35}+8=(70 - 18)\sqrt{35}+8=52\sqrt{35}+8$
When $x = -\sqrt{35}$, $f(-\sqrt{35})=2(-\sqrt{35})^{3}-18(-\sqrt{35})+8=-70\sqrt{35}+18\sqrt{35}+8=-52\sqrt{35}+8$
The points are $(\sqrt{35},52\sqrt{35}+8),(-\sqrt{35},-52\sqrt{35}+8)$

Answer:

a. $(\sqrt{3},8 - 12\sqrt{3}),(-\sqrt{3},8 + 12\sqrt{3})$
b. $(\sqrt{35},52\sqrt{35}+8),(-\sqrt{35},-52\sqrt{35}+8)$