Problem 135 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(f(x)=\left\\{\begin{array}{l}x+2,-1 \leq x<0 \\ 1, x=0 \\ \frac{x}{2}, 0

Step-by-Step Solution

Answer

(A) a minimum. (B) a maximum. (D) \(f^{\prime}(0)\) does not exist.

Step 1: Understand the Function

The function \( f(x) \) is a piecewise function defined as: - \( f(x) = x + 2 \) for \( -1 \leq x < 0 \)- \( f(x) = 1 \) at \( x = 0 \)- \( f(x) = \frac{x}{2} \) for \( 0 < x \leq 1 \)We need to evaluate this function for extrema on the interval \([-1, 1]\) and whether the derivative at 0 exists.

Step 2: Check for Extrema in Each Interval

- On \([-1, 0)\), \( f(x) = x + 2 \) is a linear function. Evaluating at endpoints, \( f(-1) = 1 \) and as \( x \to 0^- \), \( f(x) \to 2 \).- At \( x = 0 \), \( f(x) = 1 \).- On \((0, 1]\), \( f(x) = \frac{x}{2} \) is also linear. Evaluating at endpoints, \( f(0^+) = 0 \) and \( f(1) = \frac{1}{2} \).

Step 3: Compare Values to Determine Extrema

The key function values are:- \( f(-1) = 1 \)- \( f(0^-)=2 \)- \( f(0)=1 \)- \( f(0^+)=0 \)- \( f(1) = \frac{1}{2} \)- **Minimum value**: 0 at \( 0^+ \)- **Maximum value**: 2 at \( 0^- \)

Step 4: Check for Derivative Existence at \( x = 0 \)

For \( f'(0) \) to exist, the left and right derivatives at \( x = 0 \) must be equal:- Left derivative as \( x \to 0^- \) from \( f(x) = x + 2 \) is 1.- Right derivative as \( x \to 0^+ \) from \( f(x) = \frac{x}{2} \) is \( \frac{1}{2} \).Since \( 1 eq \frac{1}{2} \), \( f'(0) \) does not exist.

Key Concepts

Extrema Evaluation

In order to find the extrema of a function, it’s crucial to understand where the function achieves its minimum and maximum values over a specified interval. For the piecewise function \( f(x) \) given in segments, we're tasked with evaluating it over the interval \([-1, 1]\).

The function is broken down as follows:

• For \(-1 \leq x < 0\), \( f(x) = x + 2 \).
• At \( x = 0 \), \( f(x) = 1 \).
• For \( 0 < x \leq 1 \), \( f(x) = \frac{x}{2} \).

To find the extreme points, we assess the function values at the endpoints and any points of interest, such as where the pieces of the function meet. Specifically:

• Evaluating \( f(x) \) at the endpoint \( x = -1 \), we find \( f(-1) = 1 \).
• As \( x \to 0^- \) (approaching 0 from the left), \( f(x) \to 2 \).
• Exactly at \( x = 0 \), \( f(0) = 1 \).
• As \( x \to 0^+ \) (approaching 0 from the right), \( f(0^+) = 0 \).
• At the endpoint \( x = 1 \), \( f(1) = \frac{1}{2} \).

By comparing these values, we identify that the minimum value of the function is 0 at \( 0^+ \), and the maximum value is 2 at \( 0^- \). These provide a clear picture of the function's behavior across the interval.

Derivative Existence

Determining whether the derivative of a function exists at a given point involves checking if the function is smooth or has a pointed or corner-like turn at that point. For the given piecewise function, we need to examine the derivative at \( x = 0 \).

For the derivative \( f'(0) \) to exist, the left-hand derivative as \( x \to 0^- \) and the right-hand derivative as \( x \to 0^+ \) must be equal.
Check the left-hand derivative by considering the function as \( x \to 0^- \):

• During this approach, \( f(x) = x + 2 \). The derivative here, which is the slope of \( x + 2 \), is 1.

Then, examine the right-hand derivative for \( x \to 0^+ \):

• Here, \( f(x) = \frac{x}{2} \). The derivative in this segment is \( \frac{1}{2} \).

Since the left-hand derivative (1) does not equal the right-hand derivative (\( \frac{1}{2} \)), \( f'(0) \) does not exist. This difference indicates a sharp bend or corner at \( x = 0 \), hence no smooth transition.

Function Analysis

Analyzing a piecewise function involves looking at it both locally (within its segments) and globally (across the complete domain). The function \( f(x) \) is defined differently on three distinct intervals, each segment exhibiting unique characteristics.

Start by examining each segment:

• From \(-1 \leq x < 0\), the function \( f(x) = x + 2 \) is a linear segment, smoothly increasing.
• At \( x = 0 \), \( f(x) = 1 \) offers a constant value, representing a potential point of discontinuity with adjacent intervals.
• For \( 0 < x \leq 1 \), \( f(x) = \frac{x}{2} \) is another linear section, ascending but with a lesser slope compared to the first segment.

When piecing these components together, consider how they transform the overall shape of \( f(x) \):

• The transition at \( x = 0 \) poses a non-smooth juncture due to differing slopes on either side.
• The piecewise nature requires careful consideration of both endpoints and midpoints to fully grasp the function's behavior.
• An effective analysis covers both local attributes within individual segments and global properties that emerge when observing the entire interval \([-1, 1]\).

Piecewise function analysis blends precision in evaluating individual components with an appreciation for their collective influence on the overall mathematical landscape. This comprehensive view aids in understanding how variables interact and where extrema and derivatives might manifest or differ.