Question

An invertible function \(f(x)\) is given along with a point that lies on its graph. Using Theorem 2.7.1, evaluate \(\left(f^{-1}\right)^{\prime}(x)\) at the indicated value. \(f(x)=6 e^{3 x}\) Point \(=(0,6)\) Evaluate \(\left(f^{-1}\right)^{\prime}(6)\)

Short answer

\(\left(f^{-1}\right)'(6) = \frac{1}{18}\)

Step-by-step solution

Understand the Inverse Function Theorem

Theorem 2.7.1, likely referencing the Inverse Function Theorem, provides a formula: \[\left(f^{-1}\right)'(a) = \frac{1}{f'(f^{-1}(a))}\] where \(a\) is the point at which you need the derivative of the inverse. We need \(f(x) = a\) to find \(f^{-1}(a)\).

Investigate the Given Point

We have the point \((0, 6)\) on the graph of \(f(x) = 6e^{3x}\). This indicates that \(f(0) = 6\). Hence, \(f^{-1}(6) = 0\) because the point corresponds to \(y = f(x) = 6\) when \(x = 0\).

Differentiate the Original Function

Differentiate the given function \(f(x) = 6e^{3x}\) with respect to \(x\). Using the chain rule for differentiation, we find:\[f'(x) = 6 \cdot 3e^{3x} = 18e^{3x}\]

Substitute to Find the Derivative of the Inverse

Substitute \(f^{-1}(6) = 0\) into the derivative found in Step 3 to evaluate \(f'(0)\): \[f'(0) = 18e^{3 \cdot 0} = 18e^0 = 18\]Now use Theorem 2.7.1 to find:\[\left(f^{-1}\right)'(6) = \frac{1}{f'(0)} = \frac{1}{18}\]

Key concepts

Derivative of Inverse Function

When dealing with the Derivative of Inverse Function, it's important to remember that an inverse function, denoted as \(f^{-1}(x)\), essentially "reverses" the operation of the original function. The Inverse Function Theorem gives us a handy formula for finding the derivative of an inverse function at a particular point, which is expressed as \(\left(f^{-1}\right)'(a) = \frac{1}{f'(f^{-1}(a))}\). This means that in order to find the derivative of an inverse function at a certain value, we need to determine three things:

• The value \(a\) where we want the inverse’s derivative.
• The point that satisfies \(f(x) = a\).
• The derivative of the original function \(f\) evaluated at that specific point.

In the exercise provided, this method allows us to find \(\left(f^{-1}\right)'(6)\) by substituting our known values into the formula.

Differentiation Using Chain Rule

Differentiation using the chain rule is a critical skill when finding derivatives for functions that are compositions of other functions. The chain rule tells us how to differentiate the composition \(g(h(x))\) by working from the outside in, effectively stacking derivatives. The formula for the chain rule is \(\frac{d}{dx}g(h(x)) = g'(h(x)) \cdot h'(x)\).In the example, the function \(f(x) = 6e^{3x}\) requires the chain rule for differentiation. Here, \(e^{3x}\) is considered the "inside" function and \(6\cdot(e^{3x})\) is the "outside" function. Applying the chain rule helps us find the derivative:

• We first take the derivative of the outside function, keeping the inside function unchanged: \(6 \cdot 3e^{3x}\).
• Then, multiply by the derivative of the inside function: \(e^{3x}\) gives a derivative of \(3e^{3x}\).

See how smooth it makes the process, enabling us to easily handle exponential functions like the one presented in the exercise.

Invertible Functions

Invertible functions are essentially those which have inverses. For a function to be invertible, each output of the function must be the result of a unique input—meaning the function is one-to-one (bijective). This ensures there are no overlapping values in the output, which would prevent the reversal process necessary to define an inverse.In our particular case, the function given is \(f(x) = 6e^{3x}\). The exponential function \(e^{3x}\) inherently grows and thus satisfies the criteria of being a one-to-one function. This growth assures us that every output from \(f\) has exactly one unique pre-image in the input space.This unique mapping between input and output is what allows us to calculate \(f^{-1}(x)\)—the inverse function. Without this property, using the Inverse Function Theorem would not be feasible, because we wouldn't have the assurance needed that \(f^{-1}(x)\) exists.

Theorem Application in Calculus

Applying the Inverse Function Theorem in calculus simplifies finding derivatives of inverse functions. This theorem is a powerful tool that offers a shortcut, saving us from having to explicitly solve for the inverse function itself, which can often be complex or cumbersome.Within the context of the exercise, we are tasked to find \(\left(f^{-1}\right)'(6)\). Thanks to the theorem, once identified that \(f^{-1}(6) = 0\), we can sidestep manually inverting \(f(x) = 6e^{3x}\). Instead, we directly evaluate \[\left(f^{-1}\right)'(6) = \frac{1}{f'(0)} = \frac{1}{18}\].This immediate application of the theorem not only provides the solution efficiently but also highlights calculus's elegance in solving real-world mathematical problems. Whether dealing with exponential, polynomial, or other types of functions, the theorem's use is as versatile as it is indispensable in calculus.