Chapter 4.1, Problem 57AYU

In Problems $59-70$, for each polynomial function:

List each real zero and its multiplicity.

Determine whether the graph crosses or touches the x-axis at each x-intercept.

Determine the maximum number of turning points on the graph.

Determine the end behaviour; that is, find the power function that the graph of $f$ resembles for large values of $\left|x\right|$.

$f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

To determine

Real zeros and its multiplicity for the polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

Answer to Problem 57AYU

Solution:

The polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ has no real zeros.

Explanation of Solution

Given Information:

The polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

If $f$ is a function, then $r$ is the real solution to the equation $f\left(x\right)=0$

That is, $\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)=0$

Using the zero product rule

${\left(2x^2+9\right)}^2=0$ or $\left(x^2+7\right)=0$

To find zeros ${\left(2x^2+9\right)}^2=0$

To get rid of square, take square root on both sides

$\sqrt{{\left(2x^2+9\right)}^2}=\sqrt{0}$

$\Rightarrow \left(2x^2+9\right)=0$

Subtract 9 from both sides,

$2x^2+9-9=0-9$

$\Rightarrow 2x^2=-9$

Divide both sides by 2

$\frac{2x^2}{2}=-\frac{9}{2}$

$\Rightarrow x^2=-\frac{9}{2}$

There is no real number with its square is negative.

Thus, $2x^2+9=0$ does not have any real solution.

Now, find $\left(x^2+7\right)=0$

Subtract 7 from both sides,

$x^2+7-7=0-7$

$\Rightarrow x^2=-7$

There is no real number with its square as negative.

Thus, $x^2+7=0$ does not have any real solution.

Therefore, the polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ has no real zero.

To determine

Whether the graph of polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ crosses or touches the x-axis at each x-intercept.

Answer to Problem 57AYU

Solution:

The polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ does not cross or touches the x-axis at x-intercepts as there is no x-intercept.

Explanation of Solution

Given Information:

The polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

The x-intercepts satisfy the equation $f\left(x\right)=0$

That is, $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

From part (a), the function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ does not have any real solution.

Therefore, the polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ does not cross or touches the x-axis since there is no real zero.

To determine

The maximum number of turning points on the graph of polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

Answer to Problem 57AYU

Solution:

The maximum number of turning points on the graph of polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ is 5

Explanation of Solution

Given Information:

The polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

The polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ can be written as $f\left(x\right)=\frac{1}{2}\left(2x^2+9\right)\left(2x^2+9\right)\left(x^2+7\right)$

To simplify this polynomial, first distribute $2x^2$ inside $\left(2x^2+9\right)$ and 9 inside $\left(2x^2+9\right)$

$f\left(x\right)=\frac{1}{2}\left[\left(2x^2\left(2x^2+9\right)+9\left(2x^2+9\right)\right)\left(x^2+7\right)\right]$

By using distributive property,

$f\left(x\right)=\frac{1}{2}\left[\left(4x^4+18x^2+18x^2+81\right)\left(x^2+7\right)\right]$

Add like terms,

$f\left(x\right)=\frac{1}{2}\left[\left(4x^4+36x^2+81\right)\left(x^2+7\right)\right]$

Now, distribute $x^2$ inside $\left(4x^4+36x^2+81\right)$ and 7 inside $\left(4x^4+36x^2+81\right)$

$f\left(x\right)=\frac{1}{2}\left[\left(x^2\left(4x^4+36x^2+81\right)+7\left(4x^4+36x^2+81\right)\right)\right]$

By using distributive property,

$f\left(x\right)=\frac{1}{2}\left(4x^6+36x^4+81x^2+28x^4+252x^2+567\right)$

Add like terms,

$f\left(x\right)=\frac{1}{2}\left(4x^6+64x^4+333x^2+567\right)$

Now, distribute $\frac{1}{2}$ inside the bracket

$f\left(x\right)=2x^6+32x^4+\frac{333}{2}{x}^2+\frac{567}{2}$

This polynomial has degree 6

If $f$ is a polynomial function of degree $n$, then graph of $f$ has at most $n-1$ turning points.

Hence, the maximum number of turning points with degree 6 is $6-1=5$

Therefore, the maximum number of turning points on the graph of polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ is 5.

To determine

The end behavior that is power function that the graph of $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ resembles for large values of $\left|x\right|$

Answer to Problem 57AYU

Solution:

The end behavior that is power function that the graph of $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ resembles for large values of $\left|x\right|$ is $y=2x^6$

Explanation of Solution

Given Information:

The polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$

The end behavior of the graph of polynomial function $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\mathrm{...........}+a_{1}x+a_0{a}_n\ne 0$ is the same as that of the graph of power function $y=a_n{x}^n$

From part (c), the polynomial function $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ is written as $f\left(x\right)=2x^6+32x^4+\frac{333}{2}{x}^2+\frac{567}{2}$

Hence, the power function is $y=2x^6$

Here, the polynomial has positive leading coefficient 2 and even degree that is 6. Hence, as $x$ increases or decreases without bound $f\left(x\right)$ increases.

Therefore, the end behavior that is power function that the graph of $f\left(x\right)=\frac{1}{2}{\left(2x^2+9\right)}^2\left(x^2+7\right)$ resembles for large values of $\left|x\right|$ is, $y=2x^6$