Problem 13 - Thomas Calculus

Question

Each of Exercises \(13-18\) gives a formula for a function \(y=f(x)\) and shows the graphs of \(f\) and \(f^{-1}\) . Find a formula for \(f^{-1}\) in each case. $$ f(x)=x^{2}+1, \quad x \geq 0 $$

Step-by-Step Solution

Answer

The inverse function is \( f^{-1}(x) = \sqrt{x-1} \) for \( x \geq 1 \).

Step 1: Understand the Function and its Domain

The function given is \( f(x) = x^2 + 1 \) with the domain \( x \geq 0 \). This restriction on the domain ensures that the function is one-to-one and therefore has an inverse.

Step 2: Replace Variables in the Equation

To find the inverse, start by replacing \( y \) with \( f(x) \). The equation becomes \( y = x^2 + 1 \). Next, swap \( x \) and \( y \) to get \( x = y^2 + 1 \).

Step 3: Solve for the Inverse Function

Isolate \( y \) in the equation \( x = y^2 + 1 \). Start by subtracting 1 from both sides to get \( x - 1 = y^2 \). Then, take the square root of both sides to solve for \( y \): \( y = \sqrt{x-1} \).

Step 4: Define the Inverse Function

The inverse function \( f^{-1}(x) \) is thus \( f^{-1}(x) = \sqrt{x-1} \). Since the domain of \( f(x) \) is \( x \geq 0 \), the range of \( f^{-1}(x) \) will also start from the value \( x - 1 \geq 0 \), or \( x \geq 1 \).

Key Concepts

Function Domain

The concept of a function's domain is fundamental in understanding how functions operate. Every function has a domain, which is essentially the set of input values (or 'x' values) that the function can accept without issues. Think of it as the sandbox where the function plays.
In the case of the function \( f(x) = x^2 + 1 \), the domain is restricted to \( x \geq 0 \). This restriction is crucial because it ensures that the function is one-to-one, which means every output is matched with exactly one input. Having a one-to-one function is necessary to find its inverse.
So, when we talk about the domain of this function, we're really setting the stage to ensure we can easily flip the inputs and outputs to find the inverse function. If the domain weren't just the non-negative numbers, \( x^2 + 1 \) would produce duplicate outputs for different inputs, messing up the neat pairing needed for inversion.

Square Root Functions

Square root functions involve taking the square root of a number or expression. They can be tricky because square roots naturally have limitations based on the expression under the square root sign.
When computing the inverse of \( f(x) = x^2 + 1 \), we ended with \( f^{-1}(x) = \sqrt{x-1} \). Here, the expression under the square root, \( x-1 \), must be non-negative. This is because square roots of negative numbers are not real in basic real number arithmetic.
The domain of our inverse function, therefore, needs to be defined for \( x-1 \geq 0 \), which simplifies to \( x \geq 1 \). This ensures that we're not trying to find square roots of negative numbers, which would otherwise make the function undefined in the realm of real numbers.

Graphing Functions

Graphing functions allows us to visually interpret what's happening with different types of functions, including inverse functions. Each function has its own graphical representation based on the relationship defined between the inputs and outputs.
For \( f(x) = x^2 + 1 \), the graph is a parabola that opens upwards, starting at the point \( (0, 1) \) when \( x \) is zero. Since \( x \geq 0 \), we only look at the right side of the parabola. The inverse function \( f^{-1}(x) = \sqrt{x-1} \) then describes a square root function graph. It starts from \( (1, 0) \), rising slowly to the right.
When graphing both the function and its inverse, you often see symmetry about the line \( y = x \). This line acts as a mirror, showing how inputs have been flipped to become outputs and vice versa, a hallmark of inverse functions. Exploring graphs helps solidify the understanding of how functions and their inverses relate to each other.