Chapter 1, Problem 57EP

(a)

To determine

To find: The value of all zeros of $f$ .

Answer to Problem 57EP

There are three zeros, $x=1$ and

  $x=\frac{1\pm \sqrt{5}}{2}$

Explanation of Solution

Given information: The function is $f(x)=\frac{x^3-2x^2+1}{x^2+3}$

Calculation:

  $\frac{x^3-2x^2+1}{x^2+3}=0$

Hence the denominator is never zero, multiply by both sides by $x^2+3$

  $x^3-2x^2+1=0$

By trial and error, notice that $x=1$ is a solution.

Factor theorem: If $x=a$ is a zero of $p(x)$ then $x-a$ is a factor of $p(x)$ .

Divide $x^3-2x^2+1$ by $x-1$

So,

  $x^3-2x+1=(x-1)\left(x^2-x-1\right)$

Find the zeros of $x^2-x-1$ to use the quadratic equation:

  $\begin{array}{c}x=\frac{-(-1)\pm \sqrt{{(-1)}^2-4(1)(-1)}}{2(1)}\\ \Rightarrow x=\frac{1\pm \sqrt{1+4}}{2}\\ \Rightarrow x=\frac{1\pm \sqrt{5}}{2}\end{array}$

Therefore, there are three zeros, $x=1$ and

  $x=\frac{1\pm \sqrt{5}}{2}$ .

(b)

To determine

To find: The right end behavior model $g(x)$ for $f$ .

Answer to Problem 57EP

The right end behavior model is $g(x)=x$

Explanation of Solution

Given information: The function is $f(x)=\frac{x^3-2x^2+1}{x^2+3}$

Calculation:

As $f$ is a rational number, consider for the end behavior model the highest-degree terms of each polynomial.

Thus $g$ as:

  $\begin{array}{c}g(x)=\frac{x^3}{x^2}\\ =x\end{array}$

Therefore, the required the right end behavior model $g(x)$ for $f$ is $g(x)=x$ .

(c)

To determine

To find: The value of $\underset{x\to \infty }{\lim }f(x)$ and $\underset{x\to \infty }{\lim }\frac{f(x)}{g(x)}$

Answer to Problem 57EP

The value of

  $\begin{array}{c}\underset{x\to \infty }{\lim }f(x)=\infty \\ \underset{x\to \infty }{\lim }\frac{f(x)}{g(x)}=1\end{array}$

Explanation of Solution

Given information: The function is $f(x)=\frac{x^3-2x^2+1}{x^2+3}$

Calculation:

Know that $\underset{x\to \infty }{\lim }f(x)=\infty$ since the numerator's exponent is higher than the denominator $\underset{x\to \infty }{\lim }f(x)=\infty$

Hence, $g(x)=x$ ,get

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{f(x)}{g(x)}=\underset{x\to \infty }{\lim }\frac{\frac{x^3-2x^2+1}{x^2+3}}{x}\\ =\underset{x\to \infty }{\lim }\frac{x^3-2x^2+1}{x\cdot \left(x^2+3\right)}\\ =\underset{x\to \infty }{\lim }\frac{x^3-2x^2+1}{x^3+3x}\end{array}$

Next, multiply the highest exponent, which is $x^3$ , by the numerator and denominator.

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{x^3-2x^2+1}{x^3+3x}=\underset{x\to \infty }{\lim }\frac{\frac{x^3}{x^3}-\frac{2x^2}{x^3}+\frac{1}{x^3}}{\frac{x^3}{x^3}+\frac{3x}{x^3}}\\ =\underset{x\to \infty }{\lim }\frac{1-\frac{2}{x}+\frac{1}{x^3}}{1+\frac{3}{x^2}}\end{array}$

  $\underset{x\to \infty }{\lim }\frac{1}{x^n}=0,n>1$ So,

  $\begin{array}{c}\underset{x\to \infty }{\lim }\frac{1-\frac{2}{x}+\frac{1}{x^3}}{1+\frac{3}{x^2}}=\frac{1-0+0}{1+0}\\ =1\end{array}$

Therefore, the required value of

  $\begin{array}{c}\underset{x\to \infty }{\lim }f(x)=\infty \\ \underset{x\to \infty }{\lim }\frac{f(x)}{g(x)}=1\end{array}$