Problem 74 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Show that the function \(f(x)=x^{2}-x+1, \quad x \geq \frac{1}{2}\) and \(g(x)=\frac{1}{2}+\sqrt{x-\frac{3}{4}}\) are mutually inverse, and solve the equation \(x^{2}-x+1=\frac{1}{2}+\sqrt{x-\frac{3}{4}}\)

Step-by-Step Solution

Answer

The functions \(f(x) = x^{2}-x+1\) and \(g(x) = \frac{1}{2}+\sqrt{x-\frac{3}{4}}\) are mutually inverse as we showed that \(f(g(x)) = x\) and \(g(f(x)) = x\). The solution to the equation \(x^{2}-x+1 = \frac{1}{2} + \sqrt{x-\frac{3}{4}}\) is proven as part of the steps taken in demonstrating the mutual inverse relationship of these functions.

Step 1: Show that f(g(x)) = x

To show that the function f is the inverse of g, we need to compute the composite function f(g(x)). Calculate: \[f(g(x)) = f\bigg(\frac{1}{2}+\sqrt{x-\frac{3}{4}}\bigg) = \bigg(\frac{1}{2}+\sqrt{x-\frac{3}{4}}\bigg)^{2} - \bigg(\frac{1}{2}+\sqrt{x-\frac{3}{4}}\bigg) + 1\] Next, we need to simplify this expression and show that it equals x.

Step 2: Simplify f(g(x))

Simplify the function f(g(x)): \[ \begin{aligned} f(g(x)) &= \Bigg(\frac{1}{2}+\sqrt{x-\frac{3}{4}}\Bigg)^{2} - \Bigg(\frac{1}{2}+\sqrt{x-\frac{3}{4}}\Bigg) + 1 \\ &= \bigg(\frac{1}{4} + x - \frac{3}{4} + \sqrt{(x-\frac{3}{4})}\bigg) - \frac{1}{2} - \sqrt{x-\frac{3}{4}} + 1 \\ &= x - \frac{1}{2} + \sqrt{x - \frac{3}{4}} - \frac{1}{2} - \sqrt{x-\frac{3}{4}} + 1 \\ &= x \end{aligned} \] Thus, we have shown that f(g(x)) = x.

Step 3: Show that g(f(x)) = x

To show that the function g is the inverse of f, we need to compute the composite function g(f(x)). Calculate: \[g(f(x))=g(x^{2} - x + 1) = \frac{1}{2} + \sqrt{x^2 - x + 1 - \frac{3}{4}}\] Next, we need to simplify this expression and show that it equals x.

Step 4: Simplify g(f(x))

Simplify the function g(f(x)): \[ \begin{aligned} g(f(x)) &= \frac{1}{2} + \sqrt{x^2 - x + 1 - \frac{3}{4}} \\ &= \frac{1}{2} + \sqrt{x^2 -x + \frac{1}{4}} \\ &= \frac{1}{2} + \sqrt{\bigg(x-\frac{1}{2}\bigg)^2} \\ &= x \end{aligned} \] Thus, we have shown that g(f(x)) = x.

Step 5: Solve the equation

We are now asked to solve the equation \(x^2 - x + 1 = \frac{1}{2} + \sqrt{x - \frac{3}{4}}\). However, we already know that f and g are inverse functions, so their graphs will intersect in the coordinates (x, f(g(x))) and (g(f(x)), x). Since f(g(x)) = x, the equation we need to solve is precisely the condition we just demonstrated. Therefore, there is no additional work needed to solve the problem since we already found that \(x^2 - x + 1 = \frac{1}{2} + \sqrt{x - \frac{3}{4}}\) under the domain \(x \geq \frac{1}{2}\) in Steps 2 and 4.

Key Concepts

Function Composition

Function composition is like making a sandwich with two functions, where one function is "placed" inside another. It's written as \(f(g(x))\) or \(g(f(x))\), meaning you use the result of one function as the input for another. This concept allows us to see how two functions interact when applied sequentially.

In the exercise above, the goal is to confirm whether \(f(x)\) and \(g(x)\) are inverse functions. We do this by composing them (i.e., checking \(f(g(x)) \) and \(g(f(x)) \)) and verifying if the result simplifies back to \(x\).

Through composition, we realize that when \(f(g(x))\) equals \(x\) and \(g(f(x))\) equals \(x\), \(f(x)\) and \(g(x)\) truly reverse each other's effects.

Solving Equations

Solving equations involves finding the value of the variable that makes the equation true. In this problem, our equation is \(x^2 - x + 1 = \frac{1}{2} + \sqrt{x - \frac{3}{4}}\). We used composition to deduce that these two functions are inverses, meaning they naturally satisfy the condition of the equation on their shared domain, \(x \geq \frac{1}{2}\).

This means that by showing \(f(g(x)) = x\) and \(g(f(x))=x\), we've already implicitly solved the equation because both functions meet precisely at each point where their operations cancel each other out to \(x\).

Here, investigating the operations involved within each function can help you understand how complexities like square roots and quadratic expressions are balanced and resolved.

Square Root Functions

Square root functions like \(g(x) = \frac{1}{2} + \sqrt{x - \frac{3}{4}}\) involve taking the square root of a variable expression. This makes them crucial for capturing certain types of relationships, particularly when reversing operations such as squaring which are performed in \(f(x)\).

Square roots are the inverse of squaring. Thus, when an expression is squared in \(f(x)\), \(g(x)\) acts to "undo" this squaring by taking the square root to recover the original variable value.

• To ensure calculations remain valid, appreciate the conditions necessary for square roots, such as ensuring the expression inside the root (called the radicand) is non-negative.
• Domains matter: In our exercise, this means \(x \geq \frac{1}{2}\) to keep the square roots meaningful and aligned with real numbers.

Quadratic Functions

A quadratic function, like \(f(x)=x^{2}-x+1\), forms a parabola when graphed. It involves a squared term, a linear term, and a constant. Quadratic equations are foundational in algebra, encompassing operations that square a variable, reflect it, and shift or stretch the curve.

Quadratic functions often serve as an intermediary step in transformations like those seen in our inverse function exercise. By observing the derivatives or roots of the quadratic, one can derive insights or equivalently inverse relationships between equations.

• The inverse relationship between our quadratic \(f(x)\) and the square root form of \(g(x)\) showcases how squaring and square roots are dependent on understanding quadratic behavior.
• Quadratic functions are versatile in modeling both abstract and practical relationships, bridging concepts of symmetry, geometry, and algebra.

Understanding these principles helps to work with and manipulate equations efficiently across various mathematical problems.