Problem 269 - Calculus Volume 1

Question

Find \(\left(f^{-1}\right)^{\prime}(a)\). $$ f(x)=x^{3}+2 x+3, a=0 $$

Step-by-Step Solution

Answer

\(\left(f^{-1}\right)^{\prime}(0) = \frac{1}{5}\).

Step 1: Define the Inverse Function Derivative Formula

Given a function \( f(x) \), the derivative of its inverse at a certain point \( a \) is given by the formula: \[\left( f^{-1} \right)^{\prime}(a) = \frac{1}{f^{\prime}(f^{-1}(a))} \] Our task is to find this derivative when \( f(x) = x^3 + 2x + 3 \) and \( a = 0 \).

Step 2: Find the Derivative of f(x)

To use the inverse derivative formula, we need \( f'(x) \). Compute the derivative of \( f(x) = x^3 + 2x + 3 \):\[ f'(x) = 3x^2 + 2 \].

Step 3: Find the Value of x Such That f(x) = a

Solve \( f(x) = a \) to find \( x \). Here, we solve \[ x^3 + 2x + 3 = 0 \].Though this may seem complicated, since we need a real value of \( x \) that satisfies this relation, let's verify by trial and error. Substituting \( x = -1 \): \(-1^3 + 2(-1) + 3 = -1 -2 + 3 = 0\).So \( x = -1 \) is a solution.

Step 4: Calculate f'(f^{-1}(a))

Now that we have \( f^{-1}(0) = -1 \), we need \( f'(f^{-1}(0)) \), which equates to \( f'(-1) \).Using the derivative found previously, compute: \[ f'(-1) = 3(-1)^2 + 2 = 3(1) + 2 = 5 \].

Step 5: Compute (f^{-1})'(a)

Substitute the value into the inverse derivative formula:\[ \left( f^{-1} \right)^{\prime}(0) = \frac{1}{f^{\prime}(-1)} = \frac{1}{5} \].

Key Concepts

derivative of inverse

Understanding the derivative of an inverse function is crucial in calculus. This concept tells us how fast the inverse of a function is changing at a specific point. Contrarily to differentiating a function, this requires understanding the relationship between a function and its inverse.
The derivative of an inverse at point \( a \) is represented as \( \left( f^{-1} \right)^{\prime}(a) \). This is calculated using a special formula:

• \( \left( f^{-1} \right)^{\prime}(a) = \frac{1}{f^{\prime}(f^{-1}(a))} \)

This formula comes from the idea of swapping roles usually held by the function \( f \) and its inverse \( f^{-1} \). When calculating, it is essential to first find what value of \( x \) satisfies \( f(x)=a \). Then, the derivative \( f'(x) \) at this point is used to find the derivative of the inverse.

inverse function

An inverse function essentially undoes the work of the original function. If you plug an output value from the initial function into its inverse, it will return the original input value. Mathematically, for a function \( f(x) \), its inverse \( f^{-1}(x) \) satisfies:

• \( f(f^{-1}(x)) = x \)
• \( f^{-1}(f(x)) = x \)

Finding the inverse function involves interchanging the roles of \( x \) and \( y \) and solving for the new variable. For the function \( f(x) = x^3 + 2x + 3 \), to find its inverse, we set \( f(x) = a \) and solve for \( x \). However, finding a neat inverse expression can sometimes be complex, particularly for higher degree polynomials.

calculus problem-solving

When tackling calculus problems, a systematic approach is helpful. Identify what the problem is asking by breaking it into smaller steps. Consider using strategies like:

• Understanding the function's behavior.
• Visualizing or sketching to confirm thoughts.
• Attacking the problem step-by-step, using appropriate formulas.

For this problem, we used the derivative of an inverse function formula. It required knowing derivative basics, algebraic manipulation, and verifying our solutions with trial values. Identifying these aspects can make solving calculus problems less intimidating and more manageable.

real-world calculus applications

In real world scenarios, calculus is an indispensable tool. Calculus helps in interpreting changes, understanding growth patterns, and optimizing processes.
Functions and their inverses can model multiple real-world problems. For example, knowing how quickly a resource is being consumed requires an understanding of inverse derivatives.

• Economists use derivative concepts to evaluate marginal costs and profits.
• Engineers rely on calculus to design systems that respond to changing inputs.
• Doctors might use calculus to model the metabolic rate of drugs in the body.

These applications underline the importance of grasping the derivative of inverse functions and comprehending how functions can model and solve complex issues in various fields.