Problem 26 - Algebra and Trigonometry Real Mathematics, Real People

Question

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

Step-by-Step Solution

Answer

The function \(f(x) = \sqrt{x^{2}-1}\) is decreasing on the interval \(-\infty< x < -1\), undefined for \(-1 < x < 1\), and increasing on the interval \(1 < x < \infty\).

Step 1: Differentiate the function

By using the chain rule to differentiate the function, the derivative of the function \(f'(x)\) is found to be \[f'(x)=\frac{x}{\sqrt{x^{2}-1}}\].

Step 2: Find the critical points

Critical points occur when the derivative is zero or undefined. \[f'(x)=0\] has no solution as the fraction will never be zero. However, \(f'(x)\) is undefined at \(x = \pm 1\). So the critical points are at \(x = -1\) and \(x =1\).

Step 3: Create a number line

Using the critical points \(x= -1, 1\) to divide the real number line into sections: \(-\infty< x< -1\), \(-1 < x< 1\) and \(1< x< \infty\).

Step 4: Check each interval

Take a point in each interval and evaluate into \(f'(x)\):For \(-\infty< x< -1\), choose \(x=-2\). Substituting into \(f'(x)\) gives a negative value, hence the function is decreasing in this interval.For \(-1 < x< 1\), the function is not defined.For \(1< x< \infty\), choose \(x=2\). Substituting into \(f'(x)\) gives a positive value, so the function is increasing in this interval.