Q. 73 - Precalculus Enhanced with Graphing Utilities

Question

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

Part (b): Find the x-intercepts of the graph of f.

Part (c): Approximate any local maxima and local minima.

Part (d): Determine where f is increasing and where it is decreasing.

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

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

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

Step-by-Step Solution

Answer

Part (a): On plotting the graph, we get,

Part (b): The x-intercepts are $-3,0,3$.

Part (c): Local maximum:$10.39$ at $x=-1.73$

Local minimum: $-10.39$ at $x=1.73$

Part (d): The function is increasing in the interval $\left(-4,-1.73\right)\cup \left(1.73,4\right)$ and decreasing in the interval $\left(-1.73,1.73\right)$.

Part (e): The x-intercepts are $-5,-2,1$.

Local maximum: $10.39$ at $x=-3.73$

Local minimum: $-10.39$ at $x=-0.27$

The function is increasing in the interval $\left(-6,-3.73\right)\cup \left(-0.27,2\right)$ and decreasing in the interval $\left(-3.73,-0.27\right)$.

Part (f): The x-intercepts are $-3,0,3$.

Local maximum: $20.78$ at $x=-1.73$

Local minimum: $-20.78$ at $x=1.73$

The function is increasing in the interval $\left(-4,-1.73\right)\cup \left(1.73,4\right)$ and decreasing in the interval $\left(-1.73,1.73\right)$.

Part (g): The x-intercepts are $-3,0,3$.

Local maximum: $10.39$ at $x=1.73$

Local minimum: $-10.39$ at $x=1.73$

The function is decreasing in the interval $\left(-4,-1.73\right)\cup \left(1.73,4\right)$ and increasing in the interval $\left(-1.73,1.73\right)$.

Part (a) Step 1. Plot the function.

Consider the given function,

$f\left(x\right)=x^3-9x$ for $-4<x<4$.

Plot the graph,

Part (b) Step 1. Find the x -intercepts.

Consider the graph,

Therefore, the x-intercepts are $-3,0,3$.

Part (c) Step 1. Find the local maxima and local minima.

Consider the graph,

We can see that there is one local maxima and local minima.

Local maxima is $10.39$ at $x=-1.73$

Local minima is $-10.39$ at $x=1.73$

Part (d) Step 1. Determine where f is increasing and decreasing.

Consider the graph,

We can say that $f\left(x\right)$ is increasing in the interval $\left(-4,-1.73\right)\cup \left(1.73,4\right)$.

Also, $f\left(x\right)$ is decreasing in the interval $\left(-1.73,1.73\right)$.

Part (e) Step 1. Find the x -intercepts.

Consider the given function,

$$\begin{gathered} y=f\left(x+2\right) \\ y={\left(x+2\right)}^3-9\left(x+2\right) ...... \left(i\right) \end{gathered}$$

Substitute $y=0$ in equation (i),

$$\begin{gathered} 0={\left(x+2\right)}^3-9\left(x+2\right) \\ 0=\left\{x^3+8+6x\left(x+2\right)\right\}-9x-18 \\ 0=x^3+6x^2+3x-10 \\ 0=\left(-x+1\right)\left(x+2\right)\left(x+5\right) \\ x=1,-2,-5 \end{gathered}$$

Therefore, the x-intercepts are $-5,-2,1$.

Part (e) Step 2. Find the local maxima and local minima.

Consider the given function,

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

Local maxima and minima will have the same value but there location will change.

Hence, local maxima is $10.39$ at $x=-3.73$ and local minima is $-10.39$ at $x=-0.27$.

The function is increasing in the interval $\left(-6,-3.73\right)\cup \left(-0.27,2\right)$.

Also, the function is decreasing in the interval $\left(-3.73,-0.27\right)$.

Part (f) Step 1. Find the x -intercepts.

Consider the given function,

$$\begin{gathered} y=2f\left(x\right) \\ y=2\left(x^3-9x\right) ...... \left(i\right) \end{gathered}$$

Substitute $y=0$ in equation (i),

$$\begin{gathered} 0=2\left(x^3-9x\right) \\ 0=2x\left(x^2-9\right) \\ 0=2x\left(x+3\right)\left(x-3\right) \\ x=-3,0,3 \end{gathered}$$

Therefore, the x-intercepts are $-3,0,3$.

Part (f) Step 2. Find the local maxima and local minima.

Consider the given function,

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

Local maxima and minima will different value but there location will be same.

Hence, local maxima is $20.78$ at $x=-1.73$ and local minima is $-20.78$ at $x=1.73$.

The function is increasing in the interval $\left(-4,-1.73\right)\cup \left(1.73,4\right)$.

Also, the function is decreasing in the interval $\left(-1.73,1.73\right)$.

Part (g) Step 1. Find the x -intercepts.

Consider the given function,

$$\begin{gathered} y=f\left(-x\right) \\ y=-\left(x^3-9x\right) \\ y=-x^3+9x ...... \left(i\right) \end{gathered}$$

Substitute $y=0$ in equation (i),

$$\begin{gathered} 0=-x^3+9x \\ 0=-x\left(x^2+9\right) \\ 0=-x\left(x+3\right)\left(x-3\right) \\ x=-3,0,3 \end{gathered}$$

Therefore, the x-intercepts are $-3,0,3$.

Part (g) Step 2. Find the local maxima and local minima. Consider the given function,

Local maxima and minima will same value but there location will change.

Local maxima is $10.39$ at $x=1.73$

Local minima is $-10.39$ at $x=-1.73$

The function is decreasing in the interval $\left(-4,-1.73\right)\cup \left(1.73,4\right)$.

Also, the function is increasing in the interval $\left(-1.73,1.73\right)$.