Problem 16 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

Let \(f(x)=(x+1)^{2}-1,(x \geq-1)\), then set \(s=\left\langle x: f(x)=f^{-1}(x)\right\rangle\) is (a) Blank (b) \(<0,-1>\) (c) \(<0,1,-1>\) (d) \(\left\langle 0,-1, \frac{-3+i \sqrt{3}}{2}, \frac{-3-i \sqrt{3}}{2}\right\rangle\)

Step-by-Step Solution

Answer

The set is \( \left\langle 0, -1, \frac{-3+i\sqrt{3}}{2}, \frac{-3-i\sqrt{3}}{2} \right\rangle \).

Step 1: Understand the problem

We need to find the set of values for which the function \( f(x) \) equals its inverse \( f^{-1}(x) \). The function given is \( f(x) = (x+1)^{2} - 1 \) for \( x \geq -1 \).

Step 2: Compute the inverse function

Given \( f(x) = y = (x+1)^2 - 1 \), solve for \( x \):\[y = (x+1)^2 - 1 \Rightarrow y + 1 = (x+1)^2 \Rightarrow x+1 = \pm \sqrt{y+1}.\]Since \( x \geq -1 \), take the positive square root:\[x = \sqrt{y+1} - 1.\]Thus, the inverse function is \( f^{-1}(x) = \sqrt{x+1} - 1 \) for \( x \geq -1 \).

Step 3: Set up the equality condition

Set \( f(x) = f^{-1}(x) \):\[(x+1)^2 - 1 = \sqrt{x+1} - 1.\]Simplify to find the values of \( x \) that satisfy this equation.

Step 4: Simplify the equation

Rewriting, we get:\[(x+1)^2 - \sqrt{x+1} = 0.\]Substitute \( z = \sqrt{x+1} \), then \( z^2 = x+1 \), reducing the equation to:\[z^4 - z = 0.\]

Step 5: Solve the transformed equation

Factor the equation \( z^4 - z = 0 \):\[z(z^3 - 1) = 0.\]The solutions are \( z = 0 \) or \( z^3 = 1 \).

Step 6: Find the values of x

For \( z = 0 \), we have \( \sqrt{x+1} = 0 \), so \( x = -1 \).For \( z^3 = 1 \), the solutions are \( z = 1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \).These correspond to \( x = 0, \frac{-3+i\sqrt{3}}{2}-1, \frac{-3-i\sqrt{3}}{2}-1 \).

Step 7: Collect results in the set

The set \( s \) of solutions is \( \left\langle 0, -1, \frac{-3+i\sqrt{3}}{2}, \frac{-3-i\sqrt{3}}{2} \right\rangle \).

Key Concepts

Solving Equations

Solving equations is a fundamental skill in mathematics and involves finding the values of variables that satisfy a given equation. In this exercise, we are tasked with solving the equation where a function equals its own inverse. This involves a few strategic steps:

• First, recognize the given function, in this case, is \( f(x) = (x+1)^2 - 1 \).
• Compute its inverse, \( f^{-1}(x) \).
• Set up the equality \( f(x) = f^{-1}(x) \) and simplify.
• Transform the equation to an easier form if necessary and solve for \( x \).

To simplify the equation \((x+1)^2 - \sqrt{x+1} = 0\), we use a substitution \( z = \sqrt{x+1} \) which converts the equation to \( z^4 - z = 0 \). Solving such equations may involve factorization, as seen where \( z(z^3 - 1) = 0 \), leading to solutions for \( z \) and corresponding solutions for \( x \). Each step involves strategic simplification to make the equation more approachable.

Function Composition

Function composition involves combining two functions where the output of one function becomes the input of another. It's crucial to understand when dealing with the concept of inverse functions.

• The given problem requires finding \( f^{-1}(x) \), the inverse of \( f(x) \).
• In function composition, we say \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \), showing the function reverses its effect.
• For a function to have an inverse, it must be bijective, meaning each element in the input is mapped to a unique element in the output, and vice versa.

When composing \( f(x) = (x+1)^2 - 1 \), and finding \( f^{-1}(x) = \sqrt{x+1} - 1 \), we ensure these functions, when composed with one another, indeed result in \( x \). This ability to reverse operations showcases a precise mapping that is invertible.

Complex Numbers

Complex numbers expand the number system concept where numbers are defined with a real part and an imaginary part. Imaginary numbers denote \( i \), the square root of -1, leading to complex numbers of the form \( a + bi \).

• In this exercise's solution, the roots of the equation \( z^3 = 1 \) include complex solutions, specifically \( \frac{-1 + i\sqrt{3}}{2} \) and \( \frac{-1 - i\sqrt{3}}{2} \).
• These solutions demonstrate equations where real number solutions are insufficient.
• Complex numbers can often be found in polynomial equations involving powers, as seen in \( z^4 - z = 0 \), where roots are found beyond the real number line.

Understanding complex numbers involves recognizing how they extend solutions to equations beyond traditional real values, providing a complete set of solutions where the real line is not adequate.