Chapter 9.5, Problem 7QR

To determine

whether the function $f\left(x\right)=\frac{7x}{x^2-8}$ is both positive and decreasing on some interval $\left(N,\infty \right)$ .

Answer to Problem 7QR

Yes, the function $f\left(x\right)=\frac{7x}{x^2-8}$ is both positive and decreasing on some interval $\left(N,\infty \right)$ , where $N=2\sqrt{2}$ .

Explanation of Solution

Given information: Function $f\left(x\right)=\frac{7x}{x^2-8}$ 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 $f\left(x\right)=\frac{7x}{x^2-8}>0$ provided $x>0$ and $x^2>8$ or $x<0$ , and $x^2<8$ .

Differentiate the given function with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\frac{7x}{x^2-8}\right)\\ f^{\prime }\left(x\right)=7\frac{d}{dx}\left(\frac{x}{x^2-8}\right)\\ =7\frac{\left(x^2-8\right)\frac{d}{dx}\left(x\right)-x\frac{d}{dx}\left(x^2-8\right)}{{\left(x^2-8\right)}^2}\\ =7\frac{\left(x^2-8\right)\cdot 1-x2x}{{\left(x^2-8\right)}^2}\\ =\frac{-7x^2-56}{{\left(x^2-8\right)}^2}\end{array}$

Since for a decreasing function, $f^{\prime }\left(x\right)<0$ , $\frac{-7x^2-56}{{\left(x^2-8\right)}^2}<0$

Solve the inequality $\frac{-7x^2-56}{{\left(x^2-8\right)}^2}<0$ .

  $\begin{array}{c}\frac{-7x^2-56}{{\left(x^2-8\right)}^2}<0\\ \frac{\left(-7\left(x^2+8\right)\right)\left(-1\right)}{{\left(x+2\sqrt{2}\right)}^2{\left(x-2\sqrt{2}\right)}^2}>0\cdot \left(-1\right)\\ \frac{7\left(x^2+8\right)}{{\left(x+2\sqrt{2}\right)}^2{\left(x-2\sqrt{2}\right)}^2}>0\\ \frac{\frac{7\left(x^2+8\right)}{{\left(x+2\sqrt{2}\right)}^2{\left(x-2\sqrt{2}\right)}^2}}{7}>\frac{0}{7}\\ \frac{x^2+8}{{\left(x+2\sqrt{2}\right)}^2{\left(x-2\sqrt{2}\right)}^2}>0\end{array}$

Which gives

  $x<-2\sqrt{2}\text{ or }-2\sqrt{2}<x<2\sqrt{2}\text{ or }x>2\sqrt{2}\in$

This implies that the given function is decreasing in the interval $-\infty <x<-2\sqrt{2}$ or $--2\sqrt{2}<x<2\sqrt{2}$ or $2\sqrt{2}<x<\infty$ .

Since only the interval $\left(N,\infty \right)$ has to be considered, take only the interval $\left(2\sqrt{2},\infty \right)$ .

It is observed from the calculations that the function $f\left(x\right)=\frac{7x}{x^2-8}$ is both positive and decreasing on some interval $\left(N,\infty \right)$ .