Problem 31 - Calculus of a Single Variable

Question

Show that \(f\) is strictly monotonic on the given interval and therefore has an inverse function on that interval. \(f(x)=\frac{4}{x^{2}}, \quad(0, \infty)\)

Step-by-Step Solution

Answer

The function \(f(x)=\frac{4}{x^{2}}\) is strictly monotonic (specifically, it is strictly decreasing) on the interval \((0, \infty)\) and therefore has an inverse on that interval, given by \(f^{-1}(x) = \sqrt[3]{\frac{4}{x}}\)

Step 1: Calculate the derivative

The first step is to find the derivative of the function \(f(x)=\frac{4}{x^{2}}\). The derivative can be calculated using the power rule for derivatives, which states that the derivative of \(x^n\) is \(n*x^{n-1}\). In this case, the original function can be rewritten as \(4*x^{-2}\), so the derivative is calculated as follows: \(f'(x) = -2*4*x^{-3} = -8*x^{-3}\).

Step 2: Determine the sign of the derivative

The next step is to determine the sign of the derivative over the given interval. For \(x>0\), the derivative \(f'(x) = -8*x^{-3}\) will always be negative because a negative number times a positive number is negative.

Step 3: Determine if function is monotonic and find the inverse

Since the derivative is always negative over the interval, the function is decreasing, and therefore strictly monotonic. A function can only have an inverse if it is strictly monotonic, so we can conclude that the function does indeed have an inverse over the given interval. The inverse function of \(f(x)\) can be calculated by switching x with y and solving for y, which gives us \(f^{-1}(x) = \sqrt[3]{\frac{4}{x}}\)