Chapter 9.5, Problem 8QR

To determine

whether the function $f\left(x\right)=\frac{3+x^2}{3-x^2}$ is both positive and decreasing on some interval $\left(N,\infty \right)$ .

Answer to Problem 8QR

No , the function $f\left(x\right)=\frac{3+x^2}{3-x^2}$ is not both positive and decreasing on some interval $\left(N,\infty \right)$ .

Explanation of Solution

Given information: Function $f\left(x\right)=\frac{3+x^2}{3-x^2}$ is given.

Concept used :

If $f^{\prime }\left(x\right)<0$ , then the function is said to be decreasing.

Calculation:

It is evident from the given function that $f\left(x\right)=\frac{3+x^2}{3-x^2}>0$ provided $3-x^2>0$ or $x^2<3$ or $-\sqrt{3}<x<\sqrt{3}$ .

Differentiate the given function with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\frac{3+x^2}{3-x^2}\right)\\ f^{\prime }\left(x\right)=\frac{\left(3-x^2\right)\frac{d}{dx}\left(3+x^2\right)-\left(3+x^2\right)\frac{d}{dx}\left(3-x^2\right)}{{\left(3-x^2\right)}^2}\\ =\frac{\left(3-x^2\right)\left(2x\right)-\left(3+x^2\right)\left(-2x\right)}{{\left(x^2-8\right)}^2}\\ =\frac{6x-2x^3+6x-2x^3}{{\left(x^2-8\right)}^2}\\ =\frac{12x}{{\left(3-x^2\right)}^2}\end{array}$

Since for a decreasing function, $f^{\prime }\left(x\right)<0$ , $\frac{12x}{{\left(3-x^2\right)}^2}<0$

Solve the inequality $\frac{12x}{{\left(3-x^2\right)}^2}<0$ .

  $\begin{array}{c}\frac{12x}{{\left(3-x^2\right)}^2}<0\\ \frac{x}{{\left(3-x^2\right)}^2}<0\\ \frac{x}{{\left(x+\sqrt{3}\right)}^2{\left(x-\sqrt{3}\right)}^2}<0\\ \end{array}$

Which gives

  $x<-\sqrt{3}\text{or}-\sqrt{3}<x<0$

This implies that the given function is decreasing in the interval $-\infty <x<-\sqrt{3}$ or $-\sqrt{3}<x<0,$ .

Since only the interval $\left(N,\infty \right)$ has to be considered, none of the obtained interval lie in the given form of interval $\left(N,\infty \right)$ .

It is observed from the calculations that the function $f\left(x\right)=\frac{7x}{x^2-8}$ is positive only for $-\sqrt{3}<x<\sqrt{3}$ but not decreasing on some interval $\left(N,\infty \right)$ . As a result, there is no such interval $\left(N,\infty \right)$ , where the function is both positive and decreasing.