91Ó°ÊÓ

An inverse function essentially reverses the action of a given function. If you have a function denoted by \( f(x) \), its inverse, \( f^{-1}(x) \), will take the output of \( f(x) \) back to the original input. Here's how it works: If \( f(a) = b \), then \( f^{-1}(b) = a \). For a function to have an inverse, it must be bijective, meaning it is both one-to-one and onto. This ensures that each output is produced by exactly one input.

To find the inverse, we swap \( x \) and \( y \) in the function equation and solve for the new \( x \). In our problem, we started with \( f(x) = x^3 \) and rewrote it as \( x = y^{1/3} \) to find the inverse as \( f^{-1}(x) = x^{1/3} \). This operation effectively finds the inverse function's expression that transforms outputs back into their corresponding inputs.

The Power Rule of Differentiation is a quick and efficient method for finding the derivative of polynomial functions. If a function is given in the form \( f(x) = x^n \), its derivative \( f^{ extprime}(x) \) is calculated by multiplying the power \( n \) by \( x \) raised to the power of \( n-1 \). In other words, the derivative is \( nx^{n-1} \).

This rule simplifies the process of differentiation, especially when dealing with polynomials. For the function \( f(x) = x^3 \), the application of the power rule gives us \( f^{ extprime}(x) = 3x^2 \). This means that to find how the function rate of change varies with \( x \), you simply change the power of \( x \) by decrementing it by one, and multiply by the original power.

To differentiate an inverse function, we employ a specific formula: \( \left( f^{-1} \right)^{ extprime}(x) = \frac{1}{f^{ extprime}(f^{-1}(x))} \). This formula allows us to find the derivative of the inverse without directly differentiating the inverse function itself.

In the context of our exercise, we first find \( f^{-1}(8) \) since it leads us to \( 2 \) (because \( 2^3 = 8 \)). We then use our previously calculated \( f^{ extprime}(2) = 12 \) to evaluate:\[ \left( f^{-1} \right)^{ extprime}(8) = \frac{1}{12}. \]
This formula highlights the relationship between a function and its inverse in terms of their rates of change.

When evaluating derivatives, you find the rate at which a function is changing at a specific point. This involves substituting a specific value into the derivative function. In our case, after finding the derivative \( f^{ extprime}(x) = 3x^2 \), we evaluated it at \( x = 2 \):
\[ f^{ extprime}(2) = 3 \times 2^2 = 12. \]
This indicates that at \( x = 2 \), the original function \( f(x) = x^3 \) is increasing at a rate of 12 units per change in \( x \).

Evaluating the derivative of an inverse function at a particular point further involves using the inverse derivative formula we discussed earlier. For instance, using \( f^{ extprime}(2) \) helps find \( \left( f^{-1} \right)^{ extprime}(8) = \frac{1}{12} \). Such calculations solidify our understanding of how the inverses and derivatives interact at specific points to describe changes in behavior.