Problem 97 - Calculus for JEE ADVANCED

Question

Given that \(f(x)=(x+1)^{2}, x \geq-1\) Now, if \(g(x)\) is the reflection of \(f(x)\) in the line \(y=x\), then \(g(x)\) is an inverse function of \(y=f(x)\) Given \(y=(x+1)^{2} \quad(x \geq-1\) and \(y \geq 0)\) or \(\quad x=\pm \sqrt{y}-1\) or \(g(x)=f^{-1}(x)=\sqrt{x}-1, x \geq 0\)

Step-by-Step Solution

Answer

The inverse function is \( g(x) = \sqrt{x} - 1 \) for \( x \geq 0 \).

Step 1: Identify the Function

The function given is \( f(x) = (x+1)^2 \) for \( x \geq -1 \). This is a parabola shifted 1 unit to the left.

Step 2: Determine Conditions for Inversion

To find the inverse, we need the function to be one-to-one. Since the domain given is \( x \geq -1 \), the range of \( f(x) \) would be \( y \geq 0 \). This ensures that \( f(x) \) is one-to-one.

Step 3: Set Up the Inverse Function Equation

To find \( g(x) \), set \( y = (x+1)^2 \). To solve for \( x \), we take the square root of both sides.

Step 4: Solve for x

Isolate \( x \) by taking the square root of both sides: \[ \sqrt{y} = x + 1 \]. Thus, \( x = \sqrt{y} - 1 \). Since \( x \geq -1 \), \( y \geq 0 \) implies \( \sqrt{y} \geq 0 \), meeting both conditions.

Step 5: Write the Inverse Function

Replace \( y \) with \( x \) to express \( g(x) \) in terms of \( x \): \( g(x) = \sqrt{x} - 1 \). Since \( y \geq 0 \), the domain of \( g(x) \) is \( x \geq 0 \).

Key Concepts

Function Reflection

Function reflection revolves around the concept of reflecting a graph over a specific line. In this case, the line is the identity line, given by the equation \( y = x \). When a function is reflected over this line, its graph swaps its x and y coordinates. You can imagine folding the graph along this line, where every point on the graph of the function \( f(x) \) will now have a corresponding point at a coordinates that flip their places.

• This reflection creates the inverse function of the original function.
• For the exercise given, the function \( f(x) = (x+1)^2 \), when reflected, becomes \( f^{-1}(x) = \sqrt{x} - 1 \).

In this way, reflection over \( y = x \) provides a visual method for finding inverses, illustrating a fundamental relationship between functions and their inverses.

One-to-One Functions

A one-to-one function, or injective function, is a function in which every element of the range is mapped by exactly one element of the domain. This means no two different inputs give the same output. Understanding one-to-one functions is crucial for finding inverses.

• Before calculating an inverse, verify that the function is one-to-one. If not, the inverse won't be a function.
• In the problem, the domain \( x \geq -1 \) was chosen specifically to ensure that \( f(x) \) is one-to-one.

For parabolas, which are usually not one-to-one, restricting the domain is necessary. This makes functions like our parabola eligible for inversion.

Domain and Range

The domain refers to all possible input values for a function, while the range refers to all possible output values. In function reflections and inverses, the domain and range get swapped.

• For the original function \( f(x) = (x+1)^2 \), the domain \( x \geq -1 \) results in a range where \( y \geq 0 \).
• For the inverse function \( g(x) = \sqrt{x} - 1 \), the domain is now \( x \geq 0 \), and its range is \( y \geq -1 \).

Switching the roles of domain and range when finding an inverse helps maintain the functional structure, ensuring that every input has a unique output.

Parabolas

Parabolas are symmetric curves and are central in the discussion of certain functions and their inverses. The standard form is \( y = ax^2 + bx + c \).

• Here, the function \( f(x) = (x+1)^2 \) is a parabola shifted left by 1 unit.
• Often, parabolas pose challenges for inverses because they fail the horizontal line test, being not one-to-one.

By limiting the domain, as shown in the exercise, we make the parabola one-to-one and suitable for finding an inverse. Thus, understanding parabolas and their properties becomes essential when working with functions and their inverses.