Problem 268 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

The derivative \(\left(f^{-1}\right)'(2)\) is \(\frac{1}{3}\).

Step 1: Understand the Problem

We need to find the derivative of the inverse function \(f^{-1}\) at \(a = 2\). Given the original function \(f(x) = x^2 + 3x + 2\) for \(x \geq -1\), we first confirm that \(f\) is one-to-one and find an expression for the inverse function.

Step 2: Verify One-to-One Condition

To confirm \(f(x)\) is one-to-one, find the derivative \(f'(x) = 2x + 3\). Since \(f'(x) \geq 1\) for \(x \geq -1\), \(f(x)\) is strictly increasing and thus one-to-one on the domain \(x \geq -1\).

Step 3: Solve for the Inverse Function

To find \(f^{-1}\), set \(y = f(x) = x^2 + 3x + 2\). We need to solve for \(x\) using the quadratic formula \((\text{since the quadratic is monomial})\):\[x = \frac{-3 \pm \sqrt{3^2 - 4 \, (1) \, (2-y)}}{2 \, (1)} \] This simplifies to \[x = \frac{-3 \pm \sqrt{9 + 8 - 4y}}{2} = \frac{-3 \pm \sqrt{17 - 4y}}{2}\]. For \(x \geq -1\), the correct root is \(x = \frac{-3 + \sqrt{17 - 4y}}{2}\). Thus, \(f^{-1}(y) = \frac{-3 + \sqrt{17 - 4y}}{2}\).

Step 4: Find the Derivative of the Inverse Function

Using the formula for the derivative of the inverse function, we have:\[\left(f^{-1}\right)'(a) = \frac{1}{f'(f^{-1}(a))}\]Calculate \(f^{-1}(2)\) using the inverse function:\[f^{-1}(2) = \frac{-3 + \sqrt{17 - 8}}{2} = \frac{-3 + 3}{2} = 0\]Calculate \(f'(0)\):\[f'(x) = 2x + 3 \rightarrow f'(0) = 2 \times 0 + 3 = 3\].

Step 5: Compute \(\left(f^{-1}\right)'(2)\)

Using the formula from the previous step:\[\left(f^{-1}\right)'(2) = \frac{1}{f'(0)} = \frac{1}{3}\].

Key Concepts

Derivative of Inverse Function

Finding the derivative of an inverse function might sound challenging, but it's simpler once you break it down. The derivative of the inverse function, denoted as \(abla (f^{-1})(a)\), is related to the derivative of the original function. Inverse functions essentially "undo" what the original function does, so they have a unique relationship.

To compute the derivative of an inverse function at a point \(a\), we use the formula:

• \(abla (f^{-1})(a) = \frac{1}{f'(f^{-1}(a))}\)

This formula tells us that the derivative of the inverse at \(a\) is the reciprocal of the derivative of the original function evaluated at the inverse of \(a\). In our problem, since we know \(f'(x)\) is \(2x + 3\), we calculate it at the value \(f^{-1}(2)\). This process involves solving the function backward, which is where the inverse comes into play.

Quadratic Functions

Quadratic functions are polynomials of degree 2, generally expressed in the form \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. In our exercise, this takes the form \(f(x) = x^2 + 3x + 2\). Quadratics are graphically represented as parabolas.

A key feature of quadratic functions is their symmetry and the "U" shape of their graph. The vertex of the parabola is its peak or trough. Using the quadratic formula allows us to find the roots of the quadratic equation, which are the "solutions" or "zeroes" where the graph intersects the x-axis. This formula is used as follows:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

With a simple rearrangement, we can solve for \(x\) to find \(f^{-1}(y)\) by ensuring we choose the root appropriate for our conditions, such as \(x \geq -1\).

One-to-One Function

A function must be one-to-one (injective) to have an inverse because each output needs to come from exactly one input. This quality ensures the function's inverse is valid and unique for every input value within its domain.

To determine if a quadratic function like \(f(x) = x^2 + 3x + 2\) is one-to-one, we check its derivative. Since the derivative \(f'(x) = 2x + 3\) is positive (\(\geq 1\)) for \(x \geq -1\), it suggests that the function is always increasing. Increasing functions naturally are one-to-one because a larger input always results in a larger output compared to any preceding input.

• This monotonic behavior ensures each y-value is uniquely paired with an x-value in our domain.

Confirming that a function is one-to-one allows us to write the inverse function without conflicts or multiple values for a single input.