Problem 65 - 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 \(\left(f^{-1}\right)^{\prime}(\gamma)\). $$ f(s)=s^{3}+2 s-7, \gamma=5 $$

Step-by-Step Solution

Answer

The value of \(f^{-1}(5)\) is 2, and \((f^{-1})'(5) = \frac{1}{14}\).

Step 1: Find the Input Corresponding to Output

We need to find the value of \(s\) such that \(f(s) = 5\). This means solving the equation \(s^3 + 2s - 7 = 5\) for \(s\). First, simplify the equation: \(s^3 + 2s - 12 = 0\). Now, let's find a solution among simple integer values.

Step 2: Testing Possible Values for \(s\)

Try integer values to find \(s\) such that \(s^3 + 2s - 12 = 0\). Check with \(s = 2\): \(2^3 + 2(2) - 12 = 8 + 4 - 12 = 0\). So, \(s = 2\) is a solution and thus \(f^{-1}(5) = 2\).

Step 3: Find the Derivative of Original Function

Compute the derivative of \(f(s) = s^3 + 2s - 7\). The derivative is \(f'(s) = 3s^2 + 2\).

Step 4: Evaluate the Derivative at the found \(s\) Value

We found \(s = 2\) for \(f(s) = 5\). Now calculate \(f'(2)\): \(f'(2) = 3(2)^2 + 2 = 12 + 2 = 14\).

Step 5: Calculate the Derivative of the Inverse Function

The formula for the derivative of the inverse function is \((f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}\). Here \(b = 5\), so \((f^{-1})'(5) = \frac{1}{f'(2)} = \frac{1}{14}\).

Key Concepts

Derivative of Inverse

Understanding the concept of the derivative of an inverse function can be very useful. If you have a function \( f \) and its inverse \( f^{-1} \), the derivative of the inverse \( (f^{-1})' \) at a point \( x \) can be calculated using the formula:
\[(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\]Essentially, it tells us how fast the inverse function is changing at a particular point. Keep in mind:

• This formula relies on the derivative of \( f \) itself being non-zero at the corresponding point to avoid division by zero.
• To use it, we first need to find the point \( f^{-1}(x) \), which may involve solving equations (as discussed below).

It is fascinating because it connects the behavior of a function and its inverse, a crucial thought when dealing with inverse functions in calculus.

Cubic Function

Cubic functions are polynomials of the third degree, and they have the general form:
\[ f(x) = ax^3 + bx^2 + cx + d \]
Here, each term has a specific role:

• \( a \) dictates the steepness of the curve and whether it opens upwards or downwards.
• The terms \( b \), \( c \), and \( d \) can also influence the curve’s shape, including any bends or inflection points.

With our exercise function, \( f(s) = s^3 + 2s - 7 \), notice how the lack of a second-degree term simplifies some calculus operations. Cubic functions can have one, two, or no real roots, meaning they can curve in dynamic ways.

Solving Equations

Solving equations, especially those involving polynomials, is crucial for finding input-output relationships. To find the inverse value \( s \) for a given output \( \gamma \), you need to solve for \( s \) in the equation \( f(s) = \gamma \).
Here's the process in our exercise:

• Start by modifying the function equation to that specific instance - here \( s^3 + 2s - 7 = 5 \).
• Next, simplify the equation for ease by equating to zero: \( s^3 + 2s - 12 = 0 \).
• Trial and error with simple values is practical. Try integer values to find roots when possible, like \( s = 2 \) here.

Simple checks like substituting back into the original equation ensure the accuracy of solutions.

Function Derivative

Derivatives tell us about the rate of change of a function. For any typical function \( f(s) = s^3 + 2s - 7 \), finding the derivative involves applying power rules to each term:
\[ f'(s) = 3s^2 + 2 \]This calculation highlights:

• A derivative like \( 3s^2 + 2 \) is a polynomial derivative.
• It's used both directly for function analysis and for calculating inverses.

Evaluating the derivative at a certain point (e.g., \( s = 2 \)) aids in finding the rate at which the function value changes precisely there. Understanding derivatives is foundational for diverse applications, like physics for motion or economics for growth.