Problem 48 - Algebra for College Students

Question

(a) find \(f^{-1}\) and (b) verify that \(\left(f \circ f^{-1}\right)(x)=x\) and \(\left(f^{-1} \circ f\right)(x)=x\). $$ f(x)=x^{2}+1 \quad \text { for } x \leq 0 $$

Step-by-Step Solution

Answer

The inverse is \( f^{-1}(x) = -\sqrt{x-1} \). Both verifications hold true: \( (f \circ f^{-1})(x) = x \) and \( (f^{-1} \circ f)(x) = x \).

Step 1: Identify the Range of f

Since the function given is \( f(x) = x^2 + 1 \) for \( x \leq 0 \), this means \( x \) can be any non-positive integer. Therefore, the minimum value of \( f(x) \) is 1, making the range of \( f(x) \) as \( [1, \infty) \).

Step 2: Solving f(x) for x to find f^{-1}

To find \( f^{-1}(x) \), we substitute \( y = f(x) = x^2 + 1 \). Therefore, \( y = x^2 + 1 \). Solve for \( x \) to express it in terms of \( y \).\[y = x^2 + 1 \ y - 1 = x^2 \ x = -\sqrt{y-1}\]The negative square root is chosen because \( x \leq 0 \). Thus, \( f^{-1}(x) = -\sqrt{x - 1} \).

Step 3: Verifying (f ∘ f^{-1})(x) = x

For the verification, calculate \( (f \circ f^{-1})(x) = f(f^{-1}(x)) \).According to our results from Step 2, \( f^{-1}(x) = -\sqrt{x-1} \).Substitute into \( f(x) = x^2 + 1 \),\[ f(f^{-1}(x)) = ((-\sqrt{x-1})^2 + 1) = (x-1 + 1) = x \]Thus, \( (f \circ f^{-1})(x) = x \) is verified.

Step 4: Verifying (f^{-1} ∘ f)(x) = x

Now, verify \( (f^{-1} \circ f)(x) = f^{-1}(f(x)) \).Substitute \( f(x) = x^2 + 1 \) into \( f^{-1}(x) = -\sqrt{x - 1} \):\[f^{-1}(f(x)) = f^{-1}(x^2 + 1) = -\sqrt{x^2 + 1 - 1} = -\sqrt{x^2} = -x\]Since we defined \( f(x) \) over \( x \leq 0 \), \( -x = x \). Thus, \( (f^{-1} \circ f)(x) = x \) is verified.

Key Concepts

Function Composition

In mathematics, function composition is a fundamental concept where you apply one function to the results of another. Think of it as a two-step process, similar to baking a cake where each step depends on the previous one.
In the context of finding inverse functions, function composition helps verify if two functions are truly inverses of each other.
For any two functions, say \( f \) and \( g \), function composition is denoted as \( f \, \circ \, g \). This essentially means \( f(g(x)) \).

• When you perform \( (f \circ f^{-1})(x) \) and it results in \( x \), it indicates that \( f^{-1} \) is indeed the inverse of \( f \).
• Similarly, \( (f^{-1} \circ f)(x) = x \) provides the same confirmation from a different angle.

This dual verification through function composition is crucial. It ensures that the transformations made by applying \( f \) and then \( f^{-1} \) yield the original value of \( x \). It reassures that the inverses undo each other perfectly.

Domain and Range

Understanding the domain and range of a function is fundamental in grasping the behavior of functions, including their inverses. The domain includes all possible input values a function can accept, while the range is all possible output values it can generate.
For the given function \( f(x) = x^2 + 1 \) where \( x \leq 0 \), the domain is all non-positive real numbers. You can imagine plotting this function, starting from the highest possible negative value and moving towards zero.
The range emerges from applying the function to the domain.

• Given \( f(x) = x^2 + 1 \), even the lowest input (\( x = 0 \)) makes the function yield a minimum of 1. This means that the range is \([1, \infty)\).

The range of the original function becomes the domain of its inverse.
This swapping of domain and range is a hallmark of inverse functions, making it easier to understand and predict their behavior. Remember, the roles are reversed between a function and its inverse in these aspects.

Square Root Functions

Square root functions are critical when dealing with inverse functions, especially when the original function involves powers or squaring. They often help to 'undo' the squaring operation.
In the exercise, you start with \( f(x) = x^2 + 1 \). To find \( f^{-1}(x) \), you essentially reverse the squaring by using a square root.
Instead of directly solving for \( x \), you rearrange the formula. You isolate \( x \) when the expression is in the form of \( x^2 \), then apply the square root. To maintain consistency with the domain stipulation \( x \leq 0 \), you pick the negative of the square root:
\[ x = -\sqrt{y - 1} \] because the original function restricts \( x \) to non-positive values.
Thus, the role of the square root function is integral. It ensures the inverse function aligns with the initial conditions of the domain and range, adhering to the characteristics of these mathematical operations.