Problem 32 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(f(x)=x^{3}, \quad x \geq 1\) \(=a x+b, \quad x<1 .\) Find the constants \(a \& b\) such that \(f^{\prime}(1)\) exists. \\{Ans. \(\left.a=3, b=-2\right\\}\)

Step-by-Step Solution

Answer

The values of the constants \(a\) and \(b\) such that \(f'(1)\) exists are \(a = 3\) and \(b = -2\).

Step 1: To check continuity, we must ensure that \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)\). For \(x < 1\), \(f(x) = ax + b\). Thus, \(\lim_{x \to 1^-} f(x) = a(1) + b = a + b\). For \(x \geq 1\), \(f(x) = x^3\). Thus, \(\lim_{x \to 1^+} f(x) = 1^3 = 1\). Now, equating the limits, we have \(a + b = 1\). #Step 2: Find \(f'(x)\) for both expressions#

We need to find the derivative of the two expressions to check if they are differentiable at \(x=1\). For the expression \(ax+b\), the derivative is: \[f'(x) = \frac{d}{dx}(ax + b) = a\] For the expression \(x^3\), the derivative is: \[f'(x) = \frac{d}{dx}(x^3) = 3x^2\] #Step 3: Check the differentiability at \(x = 1\)#

Step 2: For a function to be differentiable at a point, its left and right derivatives must be equal at that point. At \(x < 1\), \(f'(x) = a\). As \(x\) approaches \(1\) from the left, \(f'(1^-) = a\). At \(x \geq 1\), \(f'(x) = 3x^2\). Thus, \(f'(1^+) = 3(1)^2 = 3\). For \(f'(1)\) to exist, \(f'(1^-) = f'(1^+)\), so \(a = 3\). #Step 4: Solve for the constants \(a\) and \(b\)#

From Step 1, we have \(a + b = 1\). From Step 3, we have \(a = 3\). Now we can find \(b\) as follows: \[b = 1 - a = 1 - 3 = -2\] #Answer#

Step 3: The values of the constants \(a\) and \(b\)

The constants \(a\) and \(b\) such that \(f'(1)\) exists are \(a = 3\) and \(b = -2\).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus that ensures a function behaves predictably without any abrupt changes across its domain. For a function to be continuous at a particular point, the limit from both directions must be equal to the function's value at that point. In our exercise, we need to check the continuity at \( x = 1 \).

• The limit from the left side (\( x \to 1^- \)) of the function is determined using the piecewise segment \( ax + b \).
• The limit from the right side (\( x \to 1^+ \)) is determined using \( x^3 \).

Equating these two limits ensures that the function does not 'jump' at \( x = 1 \). Therefore, if \( a + b = 1 \), the function remains continuous at this point.

Limits

Limits are a way to understand the behavior of functions as inputs approach a particular value. This concept is essential for determining both continuity and differentiability. In this problem, the limit serves as the tool to ensure our piecewise-defined function transitions smoothly at \( x = 1 \).
For the piecewise function \( f(x) \), to know if it is continuous at \( x = 1 \), we calculate:

• The left-hand limit: \( \lim_{x \to 1^-} f(x) = a(1) + b = a + b \).
• The right-hand limit: \( \lim_{x \to 1^+} f(x) = 1^3 = 1 \).

To satisfy both, \( a + b \) must equal \( 1 \). Only then will the limits check add up, confirming continuity at \( x = 1 \). This is crucial to establish before evaluating the derivative.

Piecewise Functions

Piecewise functions are defined by different expressions over various intervals of the input values. In our exercise, we have a function \( f(x) \) defined as:

• \( x^3 \) for \( x \geq 1 \)
• \( ax + b \) for \( x < 1 \)

The goal is to ensure these segments connect smoothly at the boundary point, \( x = 1 \). This means ensuring both continuity and differentiability. By equating the limits from both sides (ensuring \( a + b = 1 \)), we ensure there is no gap or mismatch at this point. Additionally, addressing differentiability requires consistency in the first derivatives of these segments. Thus, piecewise functions frequently require more careful analysis around the boundaries where their expressions change.

Derivatives

Derivatives are a powerful tool in calculus, providing the rate of change or the slope of a function at a given point. To ensure a function is differentiable at a certain point, its left-hand and right-hand derivatives must be equal. In this problem, we determine the derivative of each piece of our function to check differentiability at \( x = 1 \).

• For \( f(x) = ax + b \), the derivative is simply \( f'(x) = a \).
• For \( f(x) = x^3 \), the derivative is \( f'(x) = 3x^2 \), and specifically \( f'(1) = 3 \) when evaluated at \( x = 1 \).

The condition for differentiability at \( x = 1 \) is \( f'(1^-) = f'(1^+) \), meaning \( a \) must equal \( 3 \). Achieving this ensures a smooth transition with no sharp corners or discontinuities in slope at the point in question.