Chapter 1.5, Problem 35E

To determine

The function is increasing, decreasing or constant.

Answer to Problem 35E

The function is decreasing over the interval $\left(-\infty ,-1\right)$ and the graph is increasing over the interval $\left(1,\infty \right)$ .

Explanation of Solution

Given information:

The given function

Calculation:

Consider the function $f\left(x\right)=\sqrt{x^2-1}$

Graph of the equation is

A function is said to be increasing in an interval if, for any two values $x_1$ and $x_2$ in the interval,

  $x_1<x_2$ indicates $f\left(x_1\right)<f\left(x_2\right)$ .

A function is said to be decreasing in an interval if, for any two values $x_1$ and $x_2$ in the interval, $x_1<x_2$ indicates $f\left(x_1\right)>f\left(x_2\right)$ .

A function is said to be constant in an interval if, for any two values $x_1$ and $x_2$ in the interval,

  $f\left(x_1\right)=f\left(x_2\right)$

From the graph it is clear that the function is decreasing over the period $\left(-\infty ,-1\right)$ because for all points in the interval, $x_1<x_2$ indicates $f\left(x_1\right)>f\left(x_2\right)$ .

And the graph is increasing over the period $\left(1,\infty \right)$ because for all points in this interval,

  $x_1<x_2$ indicates $f\left(x_1\right)<f\left(x_2\right)$ .