91Ó°ÊÓ

The given function is $f\left(x\right)=x^2(x^2+1)(x+4)$.

Solve the given function.

$\begin{array}{rcl}f\left(x\right)& =& x^2(x^2+1)(x+4)\\ & =& x^2\left(x^3+x+4x^2+4\right)\\ & =& x^5+x^3+4x^4+4x^2\end{array}$

The polynomial function $f$is of degree $5$.

Therefore, the graph of the given function behaves like $y=x^5$.

Substitute $x=0$to find the $y$-intercepts.

$\begin{array}{rcl}f\left(x\right)& =& x^2(x^2+1)(x+4)\\ f\left(0\right)& =& 0^2(0^2+1)(0+4)\\ & =& 0\end{array}$

Thus, the $y$-intercept is at the point $(0,0)$.

Now, substitute $f\left(x\right)=0$to find the $x$-intercepts.

$\begin{array}{rcl}f\left(x\right)& =& x^2(x^2+1)(x+4)\\ 0& =& x^2(x^2+1)(x+4)\\ & ⇒& x=0\\ & & or\\ x& =& Â\pm i\\ & & or\\ x& =& -4\end{array}$

The complex numbers has no intercepts.

Thus, the $x$-intercepts are at the points $(0,0)$and $(-4,0)$.

• If the multiplicity is even, then the graph touches the $x$-axis.
• If the multiplicity is odd, then the graph crosses the $x$-axis.

From the given function $f\left(x\right)=x^2(x^2+1)(x+4)$, conclude that the zeros of the given function are $0$and width="23">-4.

The zero $0$has multiplicity $2$, so the graph of $f$touches the $x$-axis at $x=0$.

The zero $-4$has multiplicity $1$, so the graph of$f$ crosses the $x$-axis at width="51">x=-4.

The graph of the given function is shown below:

From the graph of $f$, observe that $f$has two turning points.

Using MAXIMUM, one turning point is at $(-3.17,92.15)$.

Using MINIIMUM, the other turning point is at $(0,0)$.

The graph is shown below:

Since the function is a polynomial function.

So, the domain is the set of all real numbers.

Thus, domain $(-∞,∞)$.

From the graph, the range is the set of all real numbers.

Thus, range $(-∞,∞)$.

From the graph of $f$, observe the following:

$f$is increasing on the interval $(-∞,-3.17)$and $(0,∞)$.

$f$is decreasing on the intervals $(-3.17,0)$.