Question

Self-inverse Prove that the function \(f\) is its own inverse. (a) \(f(x)=\sqrt{1-x^{2}}, \quad x \geq 0 \quad\) (b) \(f(x)=1 / x\)

Short answer

Function (a) \( f(x) = \sqrt{1 - x^2} \) is not self-inverse, but function (b) \( f(x) = \frac{1}{x} \) is self-inverse.

Step-by-step solution

Substituting Function (a) Into Itself

For function (a), perform the function on the function itself. So, replace \( x \) in \( f(x) \) with \( f(x) = \sqrt{1-x^2} \) to get \( f(f(x)) = f(\sqrt{1-x^2}) = \sqrt{1 - (1-x^2)^2} \).

Simplifying the Result of Function (a)

Simplify the expression \( \sqrt{1 - (1- x^2)^2} \) to see if it equals \( x \). Expanding the bracket yields \( \sqrt{1 - (1- 2x^2 + x^4)} \), which simplifies to \( \sqrt{1 - 1 + 2x^2 - x^4} = \sqrt{2x^2 - x^4} \). However, as we're interested in \( x \geq 0 \), we take the positive root only: \( \sqrt{x^{2}(2-x^{2})} \). Pulling out \( x^{2} \) gives \( |x| \sqrt{2-x^{2}} \). Finally, since \( x \geq 0 \) and therefore \( |x| = x \), this yields \( x \sqrt{2-x^{2}} \).

Concluding the Proof for Function (a)

As \( f(f(x)) \neq x \), function (a) is not self-inverse.

Substituting Function (b) Into Itself

For function (b), perform the function on the function itself. So, replace \( x \) in \( f(x) \) with \( f(x) = \frac{1}{x} \) to get \( f(f(x)) = f(\frac{1}{x}) = \frac{1}{\frac{1}{x}} \).

Simplifying the Result of Function (b)

Simplify \( \frac{1}{\frac{1}{x}} \) to see if it equals \( x \). By multiplicative inverse property, we get \( f(f(x)) = x \).

Concluding the Proof for Function (b)

As \( f(f(x)) = x \), we can conclude that function (b) is self-inverse.

Key concepts

Self-Inverse Functions

In calculus, a self-inverse function is a special type of function that is its own inverse. This means if you apply the function to its own output, you'll get the original input back. For instance, with function (b) from our exercise, the function defined as f(x) = 1 / x, it holds that applying f to f(x) gives back x: when we calculate f(f(x)), we are essentially taking the reciprocal of the reciprocal of x, which simplifies to x.

To prove something is a self-inverse function, we look to show that f(f(x)) = x for all x in the domain of f. This was successfully done in steps 4 and 5 of function (b). Knowing this can help students solve complex problems more quickly, as they can identify functions where this property simplifies computations.

Proof of Inverse Function

The rigorous proof of an inverse function involves showing that for every x in the domain of the function, f(f(x)) equals x. In the case of function (a), f(x) = \(\sqrt{1 - x^2}\), when we attempt to substitute f(x) into itself as shown in steps 1 to 3, we can't simplify the resultant expression to x. Instead, we get x \(\sqrt{2 - x^2}\), which is clearly not equal to x for all values of x within the domain (i.e., when x \geq 0).

This differentiation is vital because it showcases the methodology to prove that a function has an inverse, which is part of a foundational understanding in calculus. It also highlights the importance of carefully analyzing the domain of the function when considering its inverses.

Simplifying Expressions in Calculus

Calculus often requires the simplification of complex algebraic expressions to reveal properties or solutions to problems. Simplification can involve expanding brackets, factoring, or identifying patterns that aid in the reduction of the expression. For example, in the exercise, function (a) required expanding the squared binomial and simplifying the resulting expression. This process can help students move from a complex form that is difficult to analyze to a more familiar and manageable form.

Key Steps in Simplification

• Expand brackets and simplify any like terms where possible.
• Factor expressions to reveal underlying structures and patterns.
• Cancel common terms when dealing with fractions.
• Apply algebraic rules such as the multiplicative inverse property effectively, as seen in the simplification of function (b).

Understanding these steps boosts problem-solving skills and conceptual understanding, enabling students to tackle more advanced topics in calculus with confidence.