Question

let f(x)=2x^3 - 6x^2 - 90x - 3. a. identify interval(s) where the graph of f(x) is increasing. (separate intervals with commas if needed.) b. identify interval(s) where the graph of f(x) is decreasing. (separate intervals with commas if needed.) c. find all relative maximum points. (enter ordered pair(s), separated by commas if needed. write none if there is no answer.) d. find all relative minimum points. (enter ordered pair(s), separated by commas if needed. write none if there is no answer.)

Explanation:

Step1: Find the derivative of \(f(x)\)

Given \(f(x)=2x^{3}-6x^{2}-90x - 3\), using the power - rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=6x^{2}-12x - 90\).

Step2: Set \(f^\prime(x) = 0\) to find critical points

\[

$$\begin{align*} 6x^{2}-12x - 90&=0\\ x^{2}-2x - 15&=0\\ (x - 5)(x+3)&=0 \end{align*}$$

\]
The critical points are \(x=-3\) and \(x = 5\).

Step3: Determine intervals of increase and decrease

We consider the intervals \((-\infty,-3)\), \((-3,5)\) and \((5,\infty)\).

• For \(x\in(-\infty,-3)\), let's test \(x=-4\). Then \(f^\prime(-4)=6\times(-4)^{2}-12\times(-4)-90=96 + 48-90=54>0\). So \(f(x)\) is increasing on \((-\infty,-3)\).
• For \(x\in(-3,5)\), let's test \(x = 0\). Then \(f^\prime(0)=6\times0^{2}-12\times0 - 90=-90<0\). So \(f(x)\) is decreasing on \((-3,5)\).
• For \(x\in(5,\infty)\), let's test \(x = 6\). Then \(f^\prime(6)=6\times6^{2}-12\times6 - 90=216-72 - 90=54>0\). So \(f(x)\) is increasing on \((5,\infty)\).

Step4: Find relative maximum and minimum

• Since \(f(x)\) changes from increasing to decreasing at \(x=-3\), \(f(-3)=2\times(-3)^{3}-6\times(-3)^{2}-90\times(-3)-3=-54 - 54 + 270-3=159\). So the relative maximum point is \((-3,159)\).
• Since \(f(x)\) changes from decreasing to increasing at \(x = 5\), \(f(5)=2\times5^{3}-6\times5^{2}-90\times5-3=250-150 - 450-3=-353\). So the relative minimum point is \((5,-353)\).

Answer:

a. \((-\infty,-3),(5,\infty)\)
b. \((-3,5)\)
c. \((-3,159)\)
d. \((5,-353)\)