91Ó°ÊÓ

Let's rearrange the function

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

Now we have $f\left(x\right)=-x^3+4x$which has a degree of $3$Thus, the graph of $f$has an end behavior similar to $y=-x^3$

Now let's determine the intercept of $f$to get the $x-$intercept, equate $f\left(x\right)$to zero

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

So, the $x-$intercept are $x=0,x=-2$considering$f\left(0\right)=0$,thus the$y-$intercept is$(0,0)$

The zeros of $f$are $0$with the multiplicity of $1$is a xero that has an odd multiplicity, Thus the graph at crosses the x-intercept axis and $2$with the multiplicity of $1$is a xero that the has an odd multiplicity, Thus the graph crosses the $x-$axis Also, $-2$with the multiplicity of $1$is a zero that has an odd multiplicity, Thus the graph crosses the $x-$axis

Let's graph of $f$

Clearly, we can see from the graph the result gathered from step 1-2

Therefore , the function has a domain and range of all real numbers; or simply $(-∞,∞)$

By observing the graph we say that the function of $(-1.155,1.155)$is increasing; At intervals of$(-∞,1.155)$and$(1.155,∞)$is decreasing