91Ó°ÊÓ

Consider the given function,

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

Plot the graph,

Consider the graph,

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

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$x=1.15$

Consider the graph,

We can say that $f\left(x\right)$is increasing in the interval localid="1646048883892" $\left(-∞,-\sqrt{3}\right)∪\left(\sqrt{3},∞\right)$.

Also, $f\left(x\right)$is decreasing in the interval localid="1646048897956" $\left(-\sqrt{3},\sqrt{3}\right)$.

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$x=2,4,6$.

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 role="math" localid="1646050150275" $\left(-∞,2.85\right)∪\left(5.15,∞\right)$.

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

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$x=-2,0,2$.

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: role="math" localid="1646050605765" $6.16$at role="math" localid="1646050634920" $x=-1.15$

Local minima: $-6.16$at $x=1.15$

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

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

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),

role="math" localid="1646050867896" $$\begin{gathered} 0=-x^3+4x \\ 0=x\left(x^2-4\right) \\ 0=x\left(x+2\right)\left(x-2\right) \\ x=-2,0,2 \end{gathered}$$

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