Chapter 1.2, Problem 77E

Solution

(a)

To find: the smallest integer $k$ that appears to be an upper bound and graph the function.

The required value is $k=1$ .

Given information: 

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

Calculation:

The graph of $p\left(x\right)=\frac{x}{\left(1+x^2\right)}$ .

Let $y=p\left(x\right)$ .

Putting the value of $x=0$ then $y=0$ .

And $x=1$ then $y=0.5$

And $x=-1$ then $y=-0.5$

And $x=2$ then $y=0.4$

And $x=-2$ then $y=-0.4$ .

  

The smallest integer upper bound appears to be $k=1$ .

(b)

To prove: the equivalent inequality $k{x}^2-x+k>0$ .

Given information: 

The function $\frac{x}{1+x^2}<k$ .

Calculation:

Given

  $\frac{x}{1+x^2}<k$

  $1+x^2$ can never equal zero for real values of $x$ and multiply both sides by the

denominator and rearrange, obtaining

  $k{x}^2-x+k>0$

It is found $k=1$ in part $\left(a\right)$ .

Substitute:

  $x^2-x+1>0$

The discriminant:

  $\begin{array}{c}\Delta ={\left(-1\right)}^2-4\left(1\right)\left(1\right)\\ =-3\end{array}$

The discriminant is negative, the second-degree polynomial has no real zeros, and therefore the inequality is true.

(c)

To find: the greatest integer $k$ that appears to be a lower bound.

The required value is $k=-1$ .

Given information: 

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

Calculation:

The graph of $p\left(x\right)=\frac{x}{\left(1+x^2\right)}$ .

Let $y=p\left(x\right)$ .

Putting the value of $x=0$ then $y=0$ .

And $x=1$ then $y=0.5$

And $x=-1$ then $y=-0.5$

And $x=2$ then $y=0.4$

And $x=-2$ then $y=-0.4$ .

  

The greatest integer upper bound appears to be $k=-1$ .

(d)

To prove: the equivalent inequality $k{x}^2-x+k<0$ .

Given information: 

The function $\frac{x}{1+x^2}>k$ .

Calculation:

Consider the function

  $\frac{x}{1+x^2}>k$

Simplify.

  $k{x}^2-x+k<0$

Let $k=-1$ :

  $-x^2-x-1<0$

The discriminant:

  $\begin{array}{c}\Delta ={\left(-1\right)}^2-4\left(-1\right)\left(-1\right)\\ =-3\end{array}$

The discriminant is negative, the second-degree polynomial has no real zeros, and therefore the inequality is true.