Question

let f(x)=2x^3 - 6x^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)

Differentiate \(f(x)=2x^{3}-6x^{2}-18x + 8\) using the power - rule \((x^n)^\prime=nx^{n - 1}\).
\(f^\prime(x)=6x^{2}-12x-18\)

Step2: Solve for part a (horizontal tangent)

A horizontal tangent line has a slope of 0. Set \(f^\prime(x)=0\).
\[

$$\begin{align*} 6x^{2}-12x - 18&=0\\ x^{2}-2x - 3&=0\\ (x - 3)(x+1)&=0 \end{align*}$$

\]
So \(x = 3\) or \(x=-1\).
When \(x = 3\), \(f(3)=2\times3^{3}-6\times3^{2}-18\times3 + 8=2\times27-6\times9-54 + 8=54-54-54 + 8=-46\).
When \(x=-1\), \(f(-1)=2\times(-1)^{3}-6\times(-1)^{2}-18\times(-1)+8=-2-6 + 18+8=18\).
The points are \((3,-46),(-1,18)\).

Step3: Solve for part b (slope = 192)

Set \(f^\prime(x)=192\).
\[

$$\begin{align*} 6x^{2}-12x-18&=192\\ 6x^{2}-12x-210&=0\\ x^{2}-2x - 35&=0\\ (x - 7)(x + 5)&=0 \end{align*}$$

\]
So \(x = 7\) or \(x=-5\).
When \(x = 7\), \(f(7)=2\times7^{3}-6\times7^{2}-18\times7 + 8=2\times343-6\times49-126 + 8=686-294-126 + 8=274\).
When \(x=-5\), \(f(-5)=2\times(-5)^{3}-6\times(-5)^{2}-18\times(-5)+8=2\times(-125)-6\times25 + 90+8=-250-150 + 90+8=-302\).
The points are \((7,274),(-5,-302)\).

Answer:

a. \((3,-46),(-1,18)\)
b. \((7,274),(-5,-302)\)