Problem 71 - Calculus Single Variable

Question

Find \(f^{-1}(\gamma)\) for the given \(f\) and \(\gamma\) (but do not try to calculate \(f^{-1}(t)\) for a general value of \(t\) ). Then calculate \(\left(f^{-1}\right)^{\prime}(\gamma)\). $$ f(s)=\log _{2}(s)+\log _{4}(s), \gamma=9 $$

Step-by-Step Solution

Answer

\(f^{-1}(9) = 64\), \((f^{-1})'(9) = \frac{128 \ln(2)}{3}\).

Step 1: Express \(f(x)\) Using Logarithm Properties

We need to find \(f(s) = \log_{2}(s) + \log_{4}(s)\). Recall the change of base formula: \(\log_{b}(a) = \frac{\log_{c}(a)}{\log_{c}(b)}\). First, express \(\log_{4}(s)\) in terms of \(\log_{2}(s)\) by using the property that \(\log_{4}(s) = \log_{2}(s) / \log_{2}(4)\). Since \(\log_{2}(4) = 2\), we have \(\log_{4}(s) = \log_{2}(s) / 2\). Thus, \(f(s) = \log_{2}(s) + \frac{\log_{2}(s)}{2} = \frac{3}{2}\log_{2}(s)\).

Step 2: Solve for \(s\) When \(f(s) = \gamma\)

We know \(f(s) = \frac{3}{2}\log_{2}(s) = 9\). To solve for \(s\), isolate \(\log_{2}(s)\) by dividing both sides by \(\frac{3}{2}\):\[\log_{2}(s) = \frac{9 \times 2}{3} = 6\].Convert the logarithmic equation to its exponential form to solve for \(s\): \[s = 2^{6} = 64\].Thus, \(f^{-1}(9) = 64\).

Step 3: Use Inverse Function Derivative Formula

We will use the formula for the derivative of an inverse function: if \(y = f(x)\), then the derivative of the inverse function at \(y\) is given by \[(f^{-1})'(y) = \frac{1}{f'(x)}\] where \(x = f^{-1}(y)\).

Step 4: Calculate \(f'(s)\)

Given that \(f(s) = \frac{3}{2}\log_{2}(s)\), differentiate it according to the rule \(\frac{d}{dx}\log_{a}(x) = \frac{1}{x \ln(a)}\): \[f'(s) = \frac{3}{2} \times \frac{1}{s \ln(2)} = \frac{3}{2s \ln(2)}\].

Step 5: Evaluate \(f'(s)\) at \(s = 64\)

Substitute \(s = 64\) into the expression for \(f'(s)\): \[f'(64) = \frac{3}{2 \times 64 \times \ln(2)}\].

Step 6: Calculate \((f^{-1})'(9)\)

Using the result from Step 5 and the derivative of the inverse function formula from Step 3, compute:\[(f^{-1})'(9) = \frac{1}{f'(64)} = \frac{2 \times 64 \times \ln(2)}{3} = \frac{128 \ln(2)}{3}\].

Key Concepts

Logarithmic Functions

Logarithmic functions are closely tied to exponential functions, serving as their inverse. The key characteristic of a logarithm is that it answers the question: "To what power must the base be raised to produce a given number?" For example, the logarithm base 2 of 8 is 3, written as \(\log_2(8) = 3\). This is because 2 raised to the power of 3 yields 8.

Logarithms come with several properties, including:

• The product property: \(\log_b(xy) = \log_b(x) + \log_b(y)\)
• The quotient property: \(\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)\)
• The power property: \(\log_b(x^n) = n\log_b(x)\)

One widely-used property is the change of base formula, which allows us to convert logarithms from one base to another: \(\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\). This formula is particularly useful when simplifying expressions, such as converting \(\log_4(s)\) to a base of 2, making calculations more manageable.

In practice, understanding these properties can simplify complex log expressions, allowing smoother navigation through exercises that involve inverses and derivatives.

Exponential Functions

Exponential functions, expressed as \(f(x) = a^x\), depict how quantities grow or decay at consistent rates over time. The base \(a\) is always a positive constant, and the behavior of the function is largely determined by this base. When \(a > 1\), the function depicts exponential growth; if \(0 < a < 1\), it shows exponential decay.

The inverse of an exponential function is a logarithmic function. For instance, the inverse of \(y = 2^x\) is \(x = \log_2(y)\). This neat relationship between exponential and logarithmic functions is crucial for solving equations that involve these functions.

An important skill with exponentials is converting between logarithmic and exponential forms, such as transforming \(s = 2^6\) into its logarithmic counterpart \(\log_2(s) = 6\). Knowing how to manipulate these expressions helps tackle problems involving exponential growth and decay, which are common in real-world scenarios like finance, physics, and biology.

Derivatives

Derivatives are a cornerstone of calculus, measuring how a function changes as its input changes. In simple terms, the derivative at a point gives the slope of the function's graph at that point. For functions involving logarithms, the derivative follows the rule: \(\frac{d}{dx}\log_a(x) = \frac{1}{x \ln(a)}\).

In solving the inverse function problem, derivatives play a crucial role. The inverse function derivative formula states that if \(y = f(x)\), then the derivative of the inverse function at \(y\) is \((f^{-1})'(y) = \frac{1}{f'(x)}\), where \(x = f^{-1}(y)\).

• This rule helps find how steep the inverse function is at a specific point.
• This formula's application is vital in calculating \((f^{-1})'\) for specific values, guiding through steps like evaluating derivatives at exact points as shown in the exercise.

Derivatives not only define how functions behave locally but also allow us to comprehend deeper insights into their overall behavior, a fundamental aspect in understanding changes and trends in various fields, from physics to economics.