Problem 220 $$ \text { Given } f(x)=|x-1|+... [FREE SOLUTION]

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Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Sanjiva Dayal IIT Kanpur

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Chapter 4: Problem 220

$$ \text { Given } f(x)=|x-1|+|x-3|, \text { find } f^{\prime}(2) \text { . } $$

Short Answer

Expert verified

The derivative of the given function, $f(x) = |x-1| + |x-3|$, can be found by analyzing its behavior on three intervals: $x < 1$, $1 \le x < 3$, and $x \ge 3$. The derivative in each interval is: - f'(x) = -2 for $x < 1$ - f'(x) = 0 for $1 \le x < 3$ - f'(x) = 2 for $x \ge 3$ As we want to find $f'(2)$, we use the interval $1 \le x < 3$, and therefore $f'(2) = 0$.

Step by step solution

Break down the function into cases

Given the absolute value functions, we can rewrite f(x) in three different cases based on the intervals: Case 1: $x < 1$# Here, both $x - 1$ and $x - 3$ are negative. $f(x) = -(x - 1) -(x - 3)$ Case 2: $1 \le x < 3$# In this case, $x - 1$ is non-negative, while $x - 3$ is still negative. $f(x) = (x - 1) -(x - 3)$ Case 3: $x \ge 3$# Now, both $x - 1$ and $x - 3$ are non-negative. $f(x) = (x - 1) + (x - 3)$

Find the derivative for each case

Now, let's find the derivative for each case: Case 1: $x < 1$# $f'(x) = \frac{d}{dx} (-(x-1)-(x-3)) = -1 - 1 = -2$ Case 2: $1 \le x < 3$# $f'(x) = \frac{d}{dx} ((x-1)-(x-3)) = 1 - 1 = 0$ Case 3: $x \ge 3$# $f'(x) = \frac{d}{dx} ((x-1) + (x-3)) = 1 + 1 = 2$

Find the derivative at x = 2

Since we want to find $f'(2)$, we will look at the derivative in the interval where x = 2, which is Case 2: $f'(2) = 0$ So, the derivative of the function f(x) at x = 2 is 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Value Functions

Absolute value functions take any real number and return its non-negative value. They are often represented as $|x|$. In our function $f(x) = |x-1| + |x-3|$, the absolute value signs change how the function behaves depending on the value of $x$. This results in different expressions for different intervals:

For $x < 1$, both $x-1$ and $x-3$ are negative, so the absolute signs revert the signs, leading to $f(x) = -(x-1) - (x-3)$.
For $1 \le x < 3$, $x-1$ is non-negative while $x-3$ is negative, modifying the function to $f(x) = (x-1) - (x-3)$.
For $x \ge 3$, both terms are non-negative, simply giving $f(x) = (x-1) + (x-3)$.

Understanding these cases is critical when differentiating, as each interval requires a different approach. Absolute value functions can lead to piecewise-defined functions necessary for accurate calculus solutions.

Derivative of Piecewise Functions

The derivative of a piecewise function involves handling each piece or interval separately. This means you need to find the derivative for each segment according to the function's definition on each interval. From our example:

On the interval $x < 1$, the function is defined as $-(x-1)-(x-3)$. The derivative is straightforward: $f'(x) = -1 - 1 = -2$.
For $1 \le x < 3$, it's $(x-1)-(x-3)$, leading to a derivative of $f'(x) = 1 - 1 = 0$.
For $x \ge 3$, the function becomes $(x-1) + (x-3)$, and the derivative here is $f'(x) = 1 + 1 = 2$.

By understanding the piecewise definition, you can accurately determine the behavior of the derivative across the entire domain. Evaluating these separately ties back into understanding the absolute value's role in forming these intervals.

Calculus Problem Solving

In calculus problem solving, breaking down a complex problem into smaller, manageable parts is key. This procedure is particularly beneficial when dealing with piecewise or absolute value functions. Here's the general approach:

Identify the interesting points or intervals in your function. Here, the points $x=1$ and $x=3$ determine where the intervals begin and end.
Analyze the function within each interval separately. This involves writing explicit formulas without absolute values based on whether terms are positive or negative.
Find the derivative for each interval. This requires careful calculus to ensure continuity, whenever necessary.
Once derivatives are determined, evaluate at specific points to find the required value, such as $f'(2)$, known to be in the $1 \le x < 3$ interval, leading to a result of $f'(2) = 0$.

Through systematic breakdown and analysis, you can effectively tackle even complex calculus problems. This strategic method assures accuracy and a deeper understanding of mathematical behavior.

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91影视

$$ \text { Given } f(x)=|x-1|+|x-3|, \text { find } f^{\prime}(2) \text { . } $$

Short Answer

Step by step solution

Break down the function into cases

Find the derivative for each case

Find the derivative at x = 2

Key Concepts

Absolute Value Functions

Derivative of Piecewise Functions

Calculus Problem Solving

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