Problem 65 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Suppose \(f(x)=(x+1)^{2}\) for \(x \geq-1\). If \(g(x)\) is the function whose graph is the reflection of the graph of \(f(x)\) with respect to the line \(y=x\), then \(g(x)\) equals (A) \(-\sqrt{x}-1, x \geq 0\) (B) \(\frac{1}{(x+1)^{2}}, x>-1\) (C) \(\sqrt{x+1}, x \geq-1\) (D) \(\sqrt{x}-1, x \geq 0\)

Step-by-Step Solution

Answer

(D) \(\sqrt{x} - 1, x \geq 0\)

Step 1: Understand function reflection

When reflecting the graph of a function across the line \(y = x\), we swap the roles of \(x\) and \(y\). If \(y = f(x)\), the reflection would be represented by \(x = f(y)\), which gives us \(y = g(x)\) in terms of the new function.

Step 2: Start from given function

Given the original function \(f(x) = (x + 1)^2\), first we set \(y = (x + 1)^2\).

Step 3: Exchange x and y for reflection

To find the reflection of the function with respect to \(y = x\), swap \(x\) and \(y\): \(x = (y + 1)^2\).

Step 4: Solve for y

Solve the equation \(x = (y + 1)^2\) for \(y\). Taking the square root on both sides, we have \(y + 1 = \pm \sqrt{x}\). Therefore, \(y = \pm \sqrt{x} - 1\).

Step 5: Determine appropriate solution branch

Given \(x \geq -1\) for \(f(x)\), when reflecting, we need \(y = g(x)\) to have the domain \(x \geq 0\) because the range of \(f(x)\) starts from \(0\). Choose the positive branch \(y = \sqrt{x} - 1\) for a valid real solution within this reflected domain.

Step 6: Match to given options

The solution \(g(x) = \sqrt{x} - 1\) corresponds to option (D), with the condition \(x \geq 0\).

Key Concepts

Inverse Functions

An inverse function is essentially the "undoing" of a function. When we have a function \(f(x)\), it performs operations to \(x\) to give us some output \(y\). The inverse function, denoted \(f^{-1}(x)\), takes that output \(y\) and goes back to the original input \(x\). In simpler terms, if applying \(f\) to \(x\) gives \(y\), then applying \(f^{-1}\) to \(y\) gives \(x\).

• For a function to have an inverse, it must be one-to-one, meaning it passes both the vertical and horizontal line tests.
• The graphical representation of an inverse function is the reflection of the original function’s graph over the line \(y = x\).
• Consider our function \(f(x) = (x + 1)^2\). Its inverse isn’t immediately obvious due to the squaring operation, which is why solving for \(y\) by rearranging \(x = (y + 1)^2\) gives us \(y = \pm\sqrt{x} - 1\).

Thus, reflecting the function provides insight into the structure of its inverse.

Graph Transformations

Graph transformations involve changes that affect the position and shape of a graph in the coordinate plane. These modifications include translations, reflections, stretches, and compressions. Each modification alters the graph’s appearance but does not impact its fundamental characteristics.

• Translation shifts the graph horizontally or vertically.
• Reflection flips the graph over a specified line, such as the x-axis or y-axis, or even the line \(y = x\).
• Stretching and compression involve changing the graph’s dimensions without altering its overall direction.

In our exercise, the graph of \(f(x) = (x+1)^2\) is reflected over the line \(y = x\). This swaps \(x\) and \(y\) coordinates, which is why the reflection results in the form \(x = (y+1)^2\), and ultimately to \(g(x) = \sqrt{x} - 1\) after solving and choosing the correct domain.

Square Root Function

The square root function, denoted as \(\sqrt{x}\), is one of the most basic mathematical functions. It operates by finding the number that, when multiplied by itself, equals \(x\).

• The function \(y = \sqrt{x}\) has a domain of \(x \geq 0\), since taking the square root of a negative number would not produce a real number.
• It generally has a positive range, meaning \(y \geq 0\), which is paramount in function reflection problems, as seen in this exercise.

Understanding the behavior of square root functions is crucial when dealing with reflections and inverses because they often appear when solving for \(y\) in quadratic equations. In our exercise, solving for \(y\) from \(x = (y+1)^2\) led to two potential square root solutions: \(y = \sqrt{x} - 1\) and \(y = -\sqrt{x} - 1\). The correct choice depended on ensuring it fit the function's domain.