Chapter 1.2, Problem 86E

Solution

(a)

To find: the apparent horizontal asymptote of the graph.

The horizontal asymptote $1.5$ .

Given information:

The function $f\left(x\right)=\frac{3x^2-1}{2x^2+1}$ .

Calculation:

The function in the indicated viewing window.

There appears to be a horizontal asymptote at $1.5$ .

  

(b)

To find: the apparent range of $f$ .

The apparent range of $f$ is $[-1,1.5)$ .

Given information:

The function $f\left(x\right)=\frac{3x^2-1}{2x^2+1}$ .

Calculation:

Based on the graph, the range of $f\left(x\right)$ appears to be $[-1,1.5)$ .

It contains $-1$ in the range because $f\left(x\right)$ is defined for $x=-1$ .

  

(c)

To show: $-1\le \frac{3x^2-1}{2x^2+1}<1.5$ for all $x$ , thus confirming your conjecture in part (b).

Given information:

The inequality $-1\le \frac{3x^2-1}{2x^2+1}<1.5$

Calculation:

Since $2x^2+1$ is always positive (and never zero), multiply all members of the inequality by that expression:

  $\begin{array}{c}-1\left(2x^2+1\right)\le 3x^2-1<1.5\left(2x^2+1\right)\\ -2x^2-1\le 3x^2-1<3x^2+1.5\end{array}$

Add $1$ to all members of the inequality:

  $-2x^2\le 3x^2<3x^2+2.5$

Add $2x^2$

  $0\le 5x^2<5x^2+2.5$

Both parts of the inequality are true for all $x$

Note that this does not prove that $[-1,1.5)$ is the range of $f$ , but only that the range is between those two bounds.

For example, if your house is between ${23}^{\text{rd }}$ Street and ${24}^{\text{th }}$ Street, that doesn't mean that your front yard starts on one street and ends at the other.

To actually prove that range and need to use some calculus, which is beyond the scope of this textbook.

But analyze the behavior of the function for very large positive values:

  

And for very large negative values:

  

Note that as $x$ becomes very large, be it positive or negative, $f\left(x\right)$ approaches $1.5$ but never reaches that value exactly. Also note that the smallest possible value in the range is $-1$ .