Q. 74 - Precalculus Enhanced with Graphing Utilities

Question

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

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 $y=f\left(x-4\right)$.

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

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: $3.07$ at $x=-1.15$

Local minimum: $-3.07$ at $x=1.15$

Part (d): The function is increasing in the interval $\left(-\infty ,-\sqrt{3}\right)\cup \left(\sqrt{3},\infty \right)$ and decreasing in the interval $\left(-\sqrt{3},\sqrt{3}\right)$.

Part (e): The x-intercepts are $2,4,6$.

Local maximum: $3.08$ at $x=2.8$

Local minimum: $-3.08$ at $x=5.15$

The function is increasing in the interval $\left(-\infty ,2.85\right)\cup \left(5.15,\infty \right)$ and decreasing in the interval $\left(2.85,5.15\right)$.

Part (f): The x-intercepts are $x=-2,0,2$.

Local maximum: $6.16$ at $x=-1.15$

Local minimum: $-6.16$ at $x=1.15$

The function is increasing in the interval $\left(-\infty ,-1.15\right)\cup \left(1.15,\infty \right)$ and decreasing in the interval $\left(-1.15,1.15\right)$.

Part (g): The x-intercepts are $-2,0,2$.

Local maxima: $3.08$ at $x=1.15$

Local minima: $-3.08$ at $x=-1.15$

The function is increasing in the interval $\left(-1.15,1.15\right)$ and decreasing in the interval $\left(-\infty ,-1.15\right)$.

Part (a) Step 1. Plot the function.

Consider the given function,

$f\left(x\right)=x^3-4x$ for $-3<x<3$

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 $3.07$ at $x=-1.15$

Local minima is $-3.07$ at $x=1.15$

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 $\left(-\infty ,-\sqrt{3}\right)\cup \left(\sqrt{3},\infty \right)$.

Also, $f\left(x\right)$ is decreasing in the interval $\left(-\sqrt{3},\sqrt{3}\right)$.

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

Consider the given function,

$$\begin{gathered} y=f\left(x-4\right) \\ y={\left(x-4\right)}^3-4\left(x-4\right) ...... \left(i\right) \end{gathered}$$

Substitute $y=0$ in equation (i),

$$\begin{gathered} 0={\left(x-4\right)}^3-4\left(x-4\right) \\ 0={\left(x-4\right)}^3-4x+16 \\ 0=x^3-48+44x-12x^2 \\ 0=\left(x-2\right)\left(x-4\right)\left(x-6\right) \\ x=2,4,6 \end{gathered}$$

Therefore, the x-intercepts are $x=2,4,6$.

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

Consider the given function,

$y={\left(x-4\right)}^3-4\left(x-4\right)$

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

Local maxima: $3.08$ at $x=2.8$

Local minima: $-3.08$ at $x=5.15$

The function is increasing in the interval $\left(-\infty ,2.85\right)\cup \left(5.15,\infty \right)$ .

Also, the function is decreasing in the interval $\left(2.85,5.15\right)$.

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

Consider the given function,

$$\begin{gathered} y=f\left(2x\right) \\ y=2\left(x^3-4x\right) ......\left(i\right) \end{gathered}$$

Substitute $y=0$ in equation (i),

$$\begin{gathered} 0=2\left(x^3-4x\right) \\ 0=x\left(x^2-4\right) \\ 0=x\left(x+2\right)\left(x-2\right) \\ x=-2,0,2 \end{gathered}$$

Therefore, the x-intercepts are $x=-2,0,2$.

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

Consider the given function,

$y=2\left(x^3-4x\right)$

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

Local maxima: $6.16$ at $x=-1.15$

Local minima: $-6.16$ at $x=1.15$

The function is increasing in the interval $\left(-\infty ,-1.15\right)\cup \left(1.15,\infty \right)$.

Also, the function is increasing in the interval $\left(-1.15,1.15\right)$.

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

Consider the given function,

$$\begin{gathered} y=-f\left(x\right) \\ y=-\left(x^3-4x\right) \\ y=-x^3+4x ...... \left(i\right) \end{gathered}$$

Substitute $y=0$  in equation (i),

$$\begin{gathered} 0=-x^3+4x \\ 0=x\left(x^2-4\right) \\ 0=x\left(x+2\right)\left(x-2\right) \\ x=-2,0,2 \end{gathered}$$

Therefore, the x-intercepts are $-2,0,2$.

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: $3.08$ at $x=1.15$

Local minima: $-3.08$ at $x=-1.15$

The function is increasing in the interval $\left(-1.15,1.15\right)$.

Also, the function is decreasing in the interval $\left(-\infty ,-1.15\right)$.