Problem 45 - Algebra for College Students

Question

(a) find \(f^{-1}\) and (b) verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). $$ f(x)=\sqrt{x} \quad \text { for } x \geq 0 $$

Step-by-Step Solution

Answer

(a) \(f^{-1}(x)=x^2\) for \(x\geq0\). (b) Both compositions equal \(x\).

Step 1: Understand the function and its domain

The given function is \(f(x) = \sqrt{x}\), where \(x \geq 0\). This means the function takes non-negative numbers and outputs their square roots.

Step 2: Assign variables for the inverse

To find an inverse, first replace \(f(x)\) with \(y\). So, we have the equation \(y = \sqrt{x}\).

Step 3: Solve for x in terms of y

Square both sides of the equation \(y = \sqrt{x}\) to solve for \(x\). This gives us \(x = y^2\).

Step 4: Write the inverse function

Since \(x = y^2\), we replace \(y\) with \(x\) to express the inverse function: \(f^{-1}(x) = x^2\). Note that this inverse is valid for \(x \geq 0\).

Step 5: Verify \(\left(f \circ f^{-1}\right)(x) = x\)

Substitute \(f^{-1}(x)\) into \(f(x)\): \(\left(f \circ f^{-1}\right)(x) = f(x^2) = \sqrt{x^2} = x\), and is valid for \(x \geq 0\).

Step 6: Verify \(\left(f^{-1} \circ f\right)(x) = x\)

Substitute \(f(x)\) into \(f^{-1}(x)\): \(\left(f^{-1} \circ f\right)(x) = f^{-1}(\sqrt{x}) = (\sqrt{x})^2 = x\). This statement holds as long as \(x \geq 0\).

Key Concepts

Function Composition

To understand inverse functions, it's crucial to become comfortable with the idea of function composition. This means combining two functions where the output of one function becomes the input for another. An example of this is the composition \(f \circ f^{-1}\)(x), which means \(f(f^{-1}(x))\). Significantly, if \(f^{-1}\) is the true inverse of \(f\), the result of this composition should always be \(x\). The same applies for \(f^{-1} \circ f\)(x). Getting the same \(x\) back assures us that the functions are correctly inverses of each other.

• Combine functions carefully by replacing variables.
• Always check if the same variable returns.
• If \( (f \circ f^{-1})(x)\) differs from \(x\), double-check your calculations.

Domain and Range

When dealing with inverse functions, understanding the domain and range is essential. These are the potential input and output values that a function can take. For \(f(x) = \sqrt{x}\), the domain is \(x \geq 0\) because square roots of negative numbers are not real in standard mathematics.
The range of \(f\) is also \[0, \infty)\], as square roots output non-negative numbers. Note that for the inverse function \(f^{-1}(x) = x^2\), the domain is also restricted to non-negative values to ensure it is a true inverse relative to \(f\).

• A function's domain becomes the range of its inverse.
• The range of the function must accommodate possible outputs when switched as the domain of an inverse.
• Ensuring the domain and range align between a function and its inverse is crucial for proper function inversion.

Verification of Functions

Verifying that two functions are inverses involves ensuring that their compositions yield the identity function, which essentially outputs the input. To verify, you perform both compositions: \(f \circ f^{-1}\) and \(f^{-1} \circ f\). If both return \(x\), you've confirmed they're inverses.
In the given exercise:
\(f(x) = \sqrt{x}\) and its inverse is \(f^{-1}(x) = x^2\).
Verifying, \[ (f \circ f^{-1})(x) = f(x^2) = \sqrt{x^2} = x \] and \[ (f^{-1} \circ f)(x) = f^{-1}(\sqrt{x}) = (\sqrt{x})^2 = x \].
Both conditions hold true for \(x \geq 0\), confirming the functions as correct inverses.

• Always test both compositions for certainty.
• Ensure the identity \(x\) is reached in both cases.
• If discrepancies occur, re-evaluate domain, range, and calculations.