Problem 37 - Essential Calculus Early Transcendentals

Question

Find $$\left(f^{-1}\right)^{\prime}(a)$$ $$ f(x)=3+x^{2}+\tan (\pi x / 2),-1 < x < 1, \quad a=3 $$

Step-by-Step Solution

Answer

\((f^{-1})'(3) = \frac{2}{\pi}\).

Step 1: Understand the Problem

We want to find the derivative of the inverse function \(f^{-1}\) at the point \(a = 3\). The given function is \(f(x) = 3 + x^2 + \tan(\frac{\pi x}{2})\).

Step 2: Evaluate the Function at a = 3

To find \(f^{-1}(3)\), we need to determine the value of \(x\) such that \(f(x) = 3\). Substitute into the equation: \[3 + x^2 + \tan(\frac{\pi x}{2}) = 3\] Simplifying gives: \[x^2 + \tan(\frac{\pi x}{2}) = 0\].

Step 3: Solve for x

For \(x^2 + \tan(\frac{\pi x}{2}) = 0\), solve for \(x\). A simple value that satisfies this equation is \(x = 0\) because \[0^2 + \tan(0) = 0\], Thus, \(f^{-1}(3) = 0\).

Step 4: Use the Inverse Function Derivative Formula

The derivative of the inverse function \( (f^{-1})'(a) \) is found using:\[(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}\]Since \(f^{-1}(3) = 0\), this gives:\[(f^{-1})'(3) = \frac{1}{f'(0)}\].

Step 5: Compute f'(x)

Find the derivative of \(f(x) = 3 + x^2 + \tan(\frac{\pi x}{2})\):- The derivative of \(3\) is \(0\).- The derivative of \(x^2\) is \(2x\).- The derivative of \(\tan(\frac{\pi x}{2})\) is \(\frac{\pi}{2} \sec^2(\frac{\pi x}{2})\) by using the chain rule.Thus, \[f'(x) = 2x + \frac{\pi}{2} \sec^2(\frac{\pi x}{2})\].

Step 6: Evaluate f'(0)

Calculate \(f'(0)\):\[f'(0) = 2(0) + \frac{\pi}{2} \sec^2(0) = \frac{\pi}{2}\cdot 1 = \frac{\pi}{2}\].

Step 7: Substitute and Solve

Substitute \(f'(0)\) into the inverse derivative formula:\[(f^{-1})'(3) = \frac{1}{f'(0)} = \frac{1}{\frac{\pi}{2}} = \frac{2}{\pi}\].

Key Concepts

Inverse Functions

An inverse function essentially reverses the operation of the original function. If you have a function \( f(x) \) that maps an input \( x \) to an output \( y \), then its inverse \( f^{-1}(y) \) will map \( y \) back to \( x \). For instance, if \( f(x) \) adds 3 to \( x \), then \( f^{-1}(x) \) would subtract 3. In our example, given \( f(x) = 3 + x^2 + \tan\left(\frac{\pi x}{2}\right) \), finding \( f^{-1}(3) \) means identifying what \( x \) value results in \( f(x) = 3 \). In essence, if you properly understand the domain and range of your function—particularly keeping track of intervals and values involving trigonometric functions like tangent—you can identify the point where the output value is equal to the reference point, such as \( a=3 \), and find the corresponding input value, which was found to be \( x = 0 \) in this task.

Calculus Problem Solving

Solving calculus problems often involves understanding the given function and what is being asked. Inverse functions can add another layer to problem-solving since they involve reversing the outputs of the original function to determine corresponding inputs.

• Our first task was to evaluate \( f(x) \) at the point \( a = 3 \).
• We set up an equation based on the function \( f(x) = 3 + x^2 + \tan\left(\frac{\pi x}{2}\right) \).
• Simplifying this allowed us to find \( x^2 + \tan\left(\frac{\pi x}{2}\right) = 0 \).

This equation suggested checking simple values within the interval \(-1 < x < 1\). By testing \( x = 0 \), we verified that \( f(0) = 3 \), thus making \( f^{-1}(3) = 0 \). Understanding calculus problem-solving requires logical thinking and systematic approaches to manipulate equations and solve for unknown variables.

Chain Rule

The chain rule is an essential derivative rule in calculus that allows us to differentiate complex compositions of functions. It is particularly useful in situations where functions are nested within other functions. For instance, in identifying the derivative of \( \tan\left(\frac{\pi x}{2}\right) \), the chain rule assists in navigating the compound nature of the functions involved. Consider:

• The composition of functions means an outer function like \( \tan(u) \) and an inner function \( u = \frac{\pi x}{2} \).
• The chain rule tells us the derivative is \( \sec^2(u) \cdot \frac{du}{dx} \).

For \( \frac{d}{dx}\left[ \tan\left(\frac{\pi x}{2}\right) \right] \), the outer function's derivative \( \sec^2\left(\frac{\pi x}{2}\right) \) is multiplied by the derivative of the inner, \( \frac{\pi}{2} \), resulting in \( \frac{\pi}{2} \sec^2\left(\frac{\pi x}{2}\right) \). Thus, applying the chain rule facilitates obtaining the derivative of composite functions.

Derivatives

Derivatives are pivotal in calculus as they describe how functions change. In this exercise, calculating the derivative \( f'(x) \) is crucial to find \((f^{-1})'(a) \). Here are the derivative components:

• The derivative of a constant, such as \( 3 \) in \( f(x) = 3 + x^2 + \tan\left(\frac{\pi x}{2}\right) \), is zero.
• For \( x^2 \), the derivative is \( 2x \).
• Utilizing the chain rule, \( \tan\left(\frac{\pi x}{2}\right) \) has the derivative \( \frac{\pi}{2} \sec^2\left(\frac{\pi x}{2}\right) \).

Piecing these together gives us \( f'(x) = 2x + \frac{\pi}{2} \sec^2\left(\frac{\pi x}{2}\right) \). Understanding derivatives allows us to measure the rate of whatever change the function describes, leading to further analysis such as determining the slope at any function point by evaluating these derivatives.