Chapter 2.5, Problem 73AYU

67. Using a graphing utility, graph $f\left(x\right)\text{ = }{x}^3\text{ - 9}x$ for $-4\le x\le \text{ 4}$.

(a) Find the $\text{x-intercepts}$ of the graph of $f$.

(b) Approximate any local maxima and local minima.

(c) Determine where $f$ is increasing and where it is decreasing.

(d) Without using a graphing utility, repeat parts (b)-(d) for $y\text{ = }f\left(x\text{ + 2}\right)$.

(e) Without using a graphing utility, repeat parts (b)-(d) for $y\text{ = 2}f\left(x\right)$.

(f) Without using a graphing utility, repeat parts (b)-(d) for $y\text{ = }f\left(-x\right)$.

To determine

To find:

a. Using a graphing utility, graph $f\left(x\right)=x^3-9x$ for $-4\le x\le 4$.

Explanation of Solution

a. Using a graphing utility, graph $f\left(x\right)=x^3-9x$ for $-4\le x\le 4$.

To determine

To find:

b. Find the $x\text{-intercepts}$ of the graph of $f$.

Explanation of Solution

b. Find the $x\text{-intercepts}$ of the graph of $f$.

To find $x\text{-intercepts}$ put $f\left(x\right)=0$.

$f\left(x\right)=x^3-9x=0$

$x\left(x^2-9\right)=0$

$x=0\text{ and }{x}^2=9$

$x=0\text{ and }x=\pm 3$

$x=0,3,-3$

The $x\text{-intercepts}$ are $-3,0,3$.

To determine

To find:

c. Approximate any local maxima and local minima.

Explanation of Solution

c. Approximate any local maxima and local minima.

From the graph,

The function $f\left(x\right)=x^3-9x$ has $\text{local maximum}=10.392$ at $x=-1.73$.

$\text{local minimum}=-10.39$ at $x=1.73$.

To determine

To find:

d. Determine where $f$ is increasing and where it is decreasing.

Explanation of Solution

d. Determine where $f$ is increasing and where it is decreasing.

From the graph,

The function $f\left(x\right)=x^3-9x$, Increasing at $\left[-4,-1.73\right]$ and $\left[1.73,4\right]$.

Decreasing at $\left[-1.73,1.73\right]$.

To determine

To find:

e. Without using a graphing utility, repeat parts (b)–(d) for $y=f\left(x+2\right)$.

Explanation of Solution

e. (i) Find the $x\text{-intercepts}$ for $y=f\left(x+2\right)$ for $-6\le x\le 2$.

$f\left(x+2\right)={\left(x+2\right)}^3-9\left(x+2\right)$

To find $x\text{-intercepts}$ put $f\left(x\right)=0$.

$f\left(x+2\right)={\left(x+2\right)}^3-9\left(x+2\right)=0$

$\left(x+2\right)\left[{\left(x+2\right)}^2-9\right]=0$

$x=-2$ and ${\left(x+2\right)}^2\text{= }9$

$x=-2$ and ${\left(x+2\right)}^2\text{=}\pm \text{3}$

$x=-2,-5,1$

The $x\text{-intercepts}$ are $-2$, $-5$,1.

(ii) To find the local maxima and local minima:

Find the first derivative, for $f\left(x+2\right)={\left(x+2\right)}^3-9\left(x+2\right)$.

We get, $f’\left(x+2\right)=3{\left(x+2\right)}^2-9$.

Set the derivative equal to zero and solve for $x$,

$f’\left(x+2\right)=3{\left(x+2\right)}^2-9=0$

$3{\left(x+2\right)}^2=9$

${\left(x+2\right)}^2=3$

$x+2=\pm \sqrt{3}$

$x=-0.27\text{ and }x=-3.73$

Put $x=-0.27$ in $f\left(x+2\right)={\left(x+2\right)}^3-9\left(x+2\right)$,

We get, $f\left(x+2\right)={\left(-0.27+2\right)}^3-9\left(-0.27+2\right)=-10.39$.

Hence $\text{local minimum}=-10.39$ at $x=-0.27$.

Put $x=-3.73$ in $f\left(x+2\right)={\left(x+2\right)}^3-9\left(x+2\right)$,

We get, $f\left(x+2\right)={\left(-3.73+2\right)}^3-9\left(-3.73+2\right)=10.39$.

Hence $\text{local maximum}=10.39$ at $x=-3.73$.

(iii) The function $f\left(x+2\right)$ is increasing at $\left[-6,-3.73\right]$ and $\left[-0.27,2\right]$.

Decreasing at $\left[-3.73,-0.27\right]$.

To determine

To find:

f. Without using a graphing utility, repeat parts (b)–(d) for $y=2f\left(x\right)$.

Explanation of Solution

f. (i) Find the $x\text{-intercepts}$ for $y=2f\left(x\right)$ for $-4\le x\le 4$.

$2f\left(x\right)=2\left(x^3-9x\right)$

To find $x\text{-intercepts}$ put $f\left(x\right)=0$.

$2f\left(x\right)=2\left(x^3-9x\right)=0$

$2x\left(x^2-9\right)=0$

$x=0$ and $x^2=9$

$x=0$ and $x=\pm 3$

$x=0,3,-3$

The $x\text{-intercepts}$ are $-3$, 0, 3.

(ii) To find the local maxima and local minima:

Find the first derivative, for $2f\left(x\right)=2\left(x^3-9x\right)$.

We get, $2f’\left(x\right)=6x^2-18$.

Set the derivative equal to zero and solve for $x$,

$6x^2-18=0$

$6x^2=18$

$x^2=3$

$x=\pm \sqrt{3}$

$x=-1.73$ and $x=1.73$

Put $x=-1.73$ in $2f\left(x\right)=2\left(x^3-9x\right)$.

We get, $2f\left(x\right)=2\left({\left(-1.73\right)}^3–9\left(-1.73\right)\right)$.

Hence $\text{local maximum}=20.78$ at $x=-1.73$.

Put $x=1.73$ in $2f\left(x\right)=2\left(x^3-9x\right)$.

We get, $2f\left(x\right)=2\left({\left(1.73\right)}^3–9\left(1.73\right)\right)$.

Hence $\text{local minimum}=-20.78$ at $x=1.73$.

(iii) The function $2f\left(x\right)$ is increasing at $\left[-4,-1.73\right]$ and $\left[1.73,4\right]$.

Decreasing at $\left[-1.73,1.73\right]$.

To determine

To find:

g. Without using a graphing utility, repeat parts (b)–(d) for $y=f\left(-x\right)$.

Explanation of Solution

g. (i) Find the $x\text{-intercepts}$ for $y=f\left(-x\right)$ for $-4\le x\le 4$.

$f\left(-x\right)=-x^3+9x$

To find $x\text{-intercepts}$ put $f\left(x\right)=0$.

$-x^3+9x=0$

$-x\left(x^2-9\right)=0$

$-x=0$ and $x^2=9$

$x=0$ and $x=\pm 3$

$x=0,3,-3$

The $x\text{-intercepts}$ are $-3$, 0, 3.

(ii) To find the local maxima and local minima:

Find the first derivative, for $f\left(-x\right)=-x^3+9x$.

We get, $f’\left(-x\right)=-3x^2+9$.

Set the derivative equal to zero and solve for $x$,

$-3x^2+9=0$

$3x^2=9$

$x^2=3$

$x=\pm \sqrt{3}$

$x=-1.73$ and $x=1.73$

Put $x=-1.73$ in $f\left(-x\right)=-x^3+9x$.

We get, $2f\left(x\right)=-{\left(-1.73\right)}^3+9\left(-1.73\right)$.

Hence $\text{local minimum}=-10.39$ at $x=-1.73$.

Put $x=1.73$ in $f\left(-x\right)=-x^3+9x$.

We get, $f\left(-x\right)=-{\left(1.73\right)}^3+9\left(1.73\right)$.

Hence $\text{local maximum}=10.39$ at $x=1.73$.

(iii) The function $f\left(-x\right)$ is decreasing at $\left[-4,-1.73\right]$ and $\left[1.73,4\right]$.

Increasing at $\left[-1.73,1.73\right]$.