Problem 160 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The greatest value of \(f(x)=(x+1)^{1 / 3}-(x-1)^{1 / 3}\) on \([0,1]\) is: (A) 1 (B) 2 (C) 3 (D) \(1 / 3\)

Step-by-Step Solution

Answer

The greatest value is 2.

Step 1: Determine the values at the endpoints

First, evaluate \( f(x) \) at the endpoints of the interval \([0, 1]\). Compute \( f(0) \) and \( f(1) \):- For \( x = 0 \): \[ f(0) = (0+1)^{1/3} - (0-1)^{1/3} = 1 - (-1) = 2. \]- For \( x = 1 \): \[ f(1) = (1+1)^{1/3} - (1-1)^{1/3} = 2^{1/3} - 0 = 2^{1/3}. \] So, we have the values \( f(0) = 2 \) and \( f(1) = 2^{1/3} \).

Step 2: Analyze the function's behavior on the interval

Since \( f(x) = (x+1)^{1/3} - (x-1)^{1/3} \) is continuous on \([0, 1]\), any maximum would occur either at a critical point inside the interval or at one of the endpoints. We need to find if there are any critical points within the interval.Calculate the derivative \( f'(x) \):- \( f'(x) = \frac{1}{3}(x+1)^{-2/3} - \frac{1}{3}(x-1)^{-2/3} \).Set \( f'(x) = 0 \) for critical points:- \( \frac{1}{3}(x+1)^{-2/3} = \frac{1}{3}(x-1)^{-2/3} \)Simplifying gives:- \( (x+1)^{-2/3} = (x-1)^{-2/3} \)This results in no valid solution for \( x \) in \([0, 1]\).

Step 3: Compare the results from endpoints

From Step 1, we found \( f(0) = 2 \) and \( f(1) = 2^{1/3} \).Since we found that the derivative doesn't lead to any valid critical points in \([0, 1]\), the greater value must occur at one of the endpoints.Comparing \( f(0) \) and \( f(1) \):- \( 2 > 2^{1/3} \), since \( 2^{1/3} \) is approximately 1.26.Thus, the greatest value on the interval is \( f(0) = 2 \).

Key Concepts

Critical Points

When studying functions, particularly in mathematical optimization, we look for points called "critical points". Critical points are where the derivative (or slope) of a function is either zero or undefined. These points are important possible locations for local maxima, minima, or points of inflection.

To find them, we first calculate the derivative of the function. Then, we set the derivative equal to zero and solve for the values of the function variable that satisfy this equation.

• If the derivative is undefined but the function exists (like a cusp), it's also a critical point.

In the exercise, we calculated the derivative of the function and tried to find critical points by setting it to zero. However, in this case, there were no x-values within the interval \([0,1]\) that satisfied the equation, meaning no critical points existed inside the interval.

Functions

A function is a rule that assigns each input a unique output. The function \(f(x)=(x+1)^{1/3}-(x-1)^{1/3}\) relates an input \(x\) to a specific output calculated using this expression.

When dealing with functions, especially in calculus, it's important to understand how they behave over specific intervals. Functions can be continuous or have breaks and are often analyzed over certain domains such as \([0, 1]\) from our problem.

• Evaluating a function at given points can help us understand its range and limits at those endpoints.
• Continuity on the interval means the function doesn't have any breaks, making it possible to check only at endpoints and critical points.

Calculus

Calculus is a branch of mathematics that focuses on change. It provides tools like derivatives and integrals to study how functions behave and change over time.

In problems like ours, calculus helps us determine where a function reaches its maximum or minimum values, utilizing concepts like derivatives to find critical points.

• The derivative tells us about the steepness or slope of the function at any given point.
• With calculus, we can assess functions not just at endpoints, but within intervals too, though that wasn't necessary for our specific exercise as no interior critical points were found.

These concepts allow precise and clear determination of a function's behavior over intervals like \([0,1]\).

Derivatives

A derivative represents the rate at which a function is changing at any point. In simpler terms, it is the function's "slope" at a given point. Calculating derivatives is a core concept in calculus and essential for finding critical points.

For the function \(f(x)=(x+1)^{1 / 3}-(x-1)^{1 / 3}\), we calculated \(f'(x)\), which involved differentiating each term separately:

• The derivative of \( (x+1)^{1/3} \) is \( \frac{1}{3}(x+1)^{-2/3} \).
• The derivative of \( (x-1)^{1/3} \) is \( \frac{1}{3}(x-1)^{-2/3} \).

By setting the derivative \(f'(x)=0\), we explore possible critical points. As seen, solutions can sometimes indicate no critical points within a specific range, which redirected our focus to endpoints.