Problem 268 - Calculus Volume 1

Question

For each of the following functions, find \(\left(f^{-1}\right)^{\prime}(a).\) $$f(x)=x^{2}+3 x+2, x \geq-1, a=2$$

Step-by-Step Solution

Answer

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

Step 1: Verify Function is One-to-One

Since the function is defined for \(x \geq -1\), we need to check if \(f(x) = y\) is one-to-one for this domain. The quadratic function \(f(x) = x^2 + 3x + 2\) is parabolic with the vertex at \(x = -\frac{b}{2a} = -\frac{3}{2}\), which is less than \(-1\). Therefore, the function is increasing for \(x \geq -1\). Hence, \(f\) is one-to-one in the given domain.

Step 2: Find the Inverse Function

To find \(f^{-1}(x)\), solve the equation \(y = x^2 + 3x + 2\) for \(x\). Choosing \(y = a\) and rearranging, we get \(x^2 + 3x + (2 - y) = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the inverse function is \(f^{-1}(y) = -\frac{3}{2} + \frac{\sqrt{9 - 8a + 8}}{2}\).

Step 3: Differentiate the Inverse Function

By the inverse function derivative rule, \((f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}\). Differentiate \(f(x)\) to get \(f'(x) = 2x + 3\). We have \(f(x_0) = 2\) for some \(x_0\), solve \(x_0^2 + 3x_0 + 2 = 2\) giving \(x_0^2 + 3x_0 = 0\). Factor to find \(x_0(x_0 + 3) = 0\), leading to \(x_0 = 0\) or \(x_0 = -3\). Since \(x_0 \geq -1\), \(x_0 = 0\).

Step 4: Evaluate the Derivative at the Inverse

Now substitute \(x_0 = 0\) into \(f'(x)\). Compute \(f'(0) = 2(0) + 3 = 3\). Therefore, \((f^{-1})'(2) = \frac{1}{3}\).

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial that has the form \(f(x) = ax^2 + bx + c\), with \(a\), \(b\), and \(c\) as constants, and \(a eq 0\).
Quadratics are often graphed as parabolas, which can either open upwards or downwards depending on the sign of \(a\).
If \(a > 0\), the parabola opens upwards, creating a "smile" shape, whereas if \(a < 0\), it opens downwards into a "frown."

• The vertex of the parabola, a critical point, is located at \(x = -\frac{b}{2a}\).
• Quadratic functions can model various real-world situations where the rate of change is not constant.
• The solutions to the quadratic equation \(ax^2 + bx + c = 0\) can be found via factoring, completing the square, or the quadratic formula.

Studying quadratic functions helps in understanding more complex polynomial functions later on.

One-to-One Function

A one-to-one function is a special type of function where each input is related to exactly one unique output.
In other words, no two different inputs can produce the same output.
This property is crucial for a function to have an inverse that is also a function.

• To determine if a function is one-to-one, we can use the horizontal line test on its graph. If no horizontal line crosses the graph more than once, the function is one-to-one.
• Alternatively, a function \(f(x)\) is one-to-one if \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\).
• In the context of the quadratic function in this problem, restricting the domain to \(x \geq -1\) ensures it is one-to-one as it becomes an increasing function.

Recognizing one-to-one functions aids in finding their inverses and understanding their unique characteristics in mathematical analysis.

Inverse Function

The inverse of a function \(f(x)\), denoted \(f^{-1}(x)\), is a function that "reverses" the effect of \(f(x)\).
If \(f(x)\) maps \(x\) to \(y\), then \(f^{-1}(x)\) maps \(y\) back to \(x\).
This can be visualized by reflecting the graph of \(f(x)\) over the line \(y = x\).

• For a function to have an inverse, it must be one-to-one. This guarantees that the inverse is also a function.
• To find the inverse of a function, swap the roles of \(x\) and \(y\) in its equation and then solve for \(y\).
• In the provided problem, the inverse \(f^{-1}(x)\) was calculated using the quadratic formula on the equation derived from \(y = x^2 + 3x + 2\).

Mastering the concept of inverse functions is essential for solving equations and understanding the mathematical relationships that reverse one another.

Derivative of Polynomial

The derivative of a polynomial gives us the rate at which the polynomial's value changes with respect to its variable.
It is a fundamental concept in calculus, providing insights into the behavior of functions.

• For a polynomial \(f(x) = ax^n + bx^{n-1} + \ldots + c\), its derivative \(f'(x)\) is calculated by applying the power rule: \(\frac{d}{dx}[x^n] = nx^{n-1}\).
• This results in \(f'(x) = nax^{n-1} + (n-1)bx^{n-2} + \ldots\).
• In the context of the problem, the derivative \(f'(x) = 2x + 3\) was obtained for the quadratic \(f(x) = x^2 + 3x + 2\).
• Derivatives are vital in optimizing functions and understanding their increasing or decreasing nature.

Grasping the concept of polynomial derivatives enables students to tackle a wide range of calculus-related problems effectively.