Question

In Exercises \(19-30\) , find the extreme values of the function and where they occur. $$y=\frac{1}{x^{2}-1}$$

Short answer

The function \(y=\frac{1}{x^{2}-1}\) has a local maximum at x=-1, a local minimum at x=1, a global maximum at x=-1 and a global minimum at x=1.

Step-by-step solution

Find the derivative

To find extremes, we first find the derivative of the function. Apply the quotient rule for differentiation which states: (u/v)' = (v*u' - u*v')/v^2. Let \(u = 1\) and \(v = (x^2 - 1)\). Therefore, the derivative of the function, \(y'\), is: \(y' = \frac{(x^2-1)*0 - 1*2x}{(x^2-1)^2} = - \frac{2x}{(x^2-1)^2}\).

Find the Critical Numbers

Critical numbers are the x-values where the derivative is zero or undefined. Solve the equation \(y' = 0\) to determine these values. This results in no solution because the numerator of y' is not equal to zero. Then find values of x where \(y'\) is undefined which is when denominator (x^2-1)^2 equals zero. Thus, x = -1,1.

Apply the First Derivative Test

The First Derivative Test helps determine where the function reaches its local maximum and local minimum values. Find the sign of \(y'\) to the left and right of each critical number. It shows that \(y'\) changes sign from positive to negative as x goes from less than -1 to greater than -1. Hence, x = -1 is a local maximum. Also, \(y'\) changes sign from negative to positive as x goes from less than 1 to greater than 1 and hence, x = 1 is a local minimum.

Find Global Extremes

To find global extremes, examine the end behaviour of the function. The function approaches 0 as \(x \rightarrow -\infty\) and \(x \rightarrow \infty\). Using limits, \[ \lim_{x \rightarrow \pm \infty} \frac{1}{x^2-1} = 0\] Also as x approaches critical values, \(y \rightarrow -\infty\) for x = -1 and \(y \rightarrow \infty\) for x = 1. Thus, the global maximum exists at x = -1 and global minimum exists at x = 1.