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Chapter 1: Problem 42

Determine the intervals over which the function is increasing, decreasing, or constant. $$ f(x)=\sqrt{x^{2}-1} $$

Short Answer

Expert verified
The function \(f(x) = \sqrt{x^{2}-1}\) is decreasing on the interval \((-鈭�, -1)\), increasing on the interval \((1, 鈭�)\), and is not defined in the interval \((-1, 1)\). The function is not constant at any interval.

Step by step solution

01

Differentiate the function

The first step is to differentiate the function \(f(x) = \sqrt{x^{2}-1}\) with respect to \(x\) to attain its derivative. This can be accomplished by using the Chain Rule. The derivative is \(f'(x) = \frac{x}{\sqrt{x^{2}-1}}\).
02

Find critical points

The second step is to set the derivative to zero or undefined and solve the equation. A critical point occurs where the derivative of a function is zero or undefined. Because the denominator of \(f'(x)\) can never be zero, there are no critical points from \(f'(x) = 0\). However, there is one critical point from the undefined values, at \(x = 卤1\).
03

Interval analysis

Next is to do an interval analysis. Choose test points to the left and right of each critical point and substitute them into the derivative function. If the result is positive, the function is increasing at that interval, if negative it's decreasing, and if it's equal to zero the function is constant. For example, choose -2, 0, and 2 as test points. Plugging these values into \(f'(x)\), we get \(f'(-2) < 0\), \(f'(0)\) is undefined, and \(f'(2) > 0\). This means that the function is decreasing on the interval \((-鈭�, -1)\), and increasing on the interval \((1, 鈭�)\). It's undefined on the interval \((-1, 1)\), because \(\sqrt{x^{2}-1}\) is not defined for \(|x| < 1\).
04

Discussion about the constant part

The function isn't a constant function because the value of the function continuously changes with respect to x. Thus, there are no intervals where the function is constant.

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