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Chapter 13: Problem 15

If \(f(x)=|x-1|\) and \(g(x)=f\\{f[f(x)]\\}\), then for \(x>2\), \(g^{\prime}(x)\) is equal to (A) \(-1\) if \(2 \leq x<3\) (B) 1 if \(2 \leq x<3\) (C) 1 for all \(x>2\) (D) None of these

Short Answer

Expert verified
For \( x > 2 \), \( g'(x) = 1 \), so (C) is correct.

Step by step solution

01

Understand the Function Definitions

We start by analyzing the given functions. The function \( f(x) = |x-1| \) represents the absolute value function, which returns the positive difference between \( x \) and 1. This means \( f(x) = x - 1 \) for \( x \geq 1 \) and \( f(x) = 1 - x \) for \( x < 1 \). For \( g(x) = f\{f[f(x)]\} \), it indicates the repeated application of the absolute function three times.
02

Evaluate f(x) for x > 2

For \( x > 2 \), we have \( f(x) = |x - 1| = x - 1 \) since \( x \geq 1 \), ensuring that the expression inside the absolute value is non-negative. Hence, when \( x > 2 \), simplifying \( f(x) \) gives \( x - 1 \).
03

Evaluate f(f(x)) for x > 2

Given \( f(x) = x - 1 \) for \( x > 2 \), we evaluate \( f(f(x)) = f(x - 1) = |(x - 1) - 1| = |x - 2| \). Because \( x > 2 \), we have \( x - 2 > 0 \) and thus, \( f(f(x)) = x - 2 \).
04

Evaluate f(f(f(x))) for x > 2

Next, we find \( f(f(f(x))) = f(x - 2) \). This results in \( |x - 2 - 1| = |x - 3| \). For \( x > 2 \), \( x - 3 \) could be negative or positive, but since \( x > 3 \), it will be positive: \( x - 3 \). Hence, \( f(f(f(x))) = x - 3 \) for \( x > 3 \).
05

Differentiate g(x) with respect to x for x > 3

Next, we differentiate \( g(x) = x - 3 \) with respect to \( x \). The derivative of a linear function \( ax + b \) is simply \( a \). So, \( g'(x) = 1 \).
06

Conclusion on g'(x) for x > 2

Given the piecewise nature of the differentiation depending on \( x \), for \( x > 3 \), we found \( g'(x) = 1 \). Therefore, for \( x > 2 \) with \( x > 3 \) in focus, the answer aligns most closely with (C) since it's constant and doesn't switch signs within \( x > 2 \), but we focus here on where the differentiable portion applies.

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Key Concepts

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

Absolute Value Function
The absolute value function is a fundamental concept in calculus and mathematics. It represents the distance a number is from zero on the number line, ignoring its direction. The notation for absolute value is expressed as \(|x|\), where \(x\) can be any real number.
When working with absolute values, consider the following properties:
  • If \(x\) is positive or zero, \(|x| = x\).
  • If \(x\) is negative, \(|x| = -x\).
In the context of our exercise, we have the function \(f(x) = |x - 1|\). This equation measures the distance of \(x\) from 1.
Depending on the input \(x\):
  • If \(x \geq 1\), then \(f(x) = x - 1\).
  • If \(x < 1\), then \(f(x) = 1 - x\).
These two pieces make the function a piecewise function, which leads us to our next core concept.
Piecewise Function
A piecewise function is one that is defined using different expressions based on the input value domain. In simpler terms, it consists of "pieces" of different functions. Each piece applies to a specific interval of input values.
To illustrate with our exercise:
  • For \(f(x) = |x - 1|\), the function is defined by two linear expressions depending on whether \(x\) is greater than or equal to 1, or less than 1.
When examining more complex applications of piecewise functions like in the composite function \(g(x) = f(f(f(x)))\), it's important to:
  • Apply each piece of the function based on the interval of \(x\).
  • Reevaluate the sub-functions using piecewise definitions at each step.
For example, in our exercise, we determined \(f(f(f(x))) = x - 3\) for \(x > 3\). Piecewise functions, therefore, demand attention at each transition in the domain where the expression for the function changes.
Differentiation
Differentiation is a core concept in calculus, focusing on finding the rate at which function values change concerning changes in their input. Simply put, it helps us understand how the output of a function grows or diminishes as its input moves.
To differentiate a function manually, you apply derivative rules, such as:
  • The power rule: If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).
  • The constant rule: The derivative of a constant is zero.
  • The sum rule: The derivative of the sum is the sum of the derivatives.
In our exercise, we sought to find \(g'(x)\) from the function \(g(x) = x - 3\) for \(x > 3\). Given the linear form of this function as \(x - 3\), the derivative is straightforward:
  • The derivative \(g'(x)\) is 1, since the slope of the linear function is the coefficient of \(x\).
Thus, understanding differentiation reveals how functions behave dynamically, crucial for unraveling the mechanics of calculus-based problems.

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Most popular questions from this chapter

Instructions: In the following questions an Assertion (A) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: If \(f(x)=(\cos x+i \sin x)(\cos 2 x+i \sin 2 x)\) \((\cos 3 x+i \sin 3 x) \ldots(\cos n x+i \sin n x)\) and \(f(1)=1\) then \(f^{\prime \prime}(1)\) is equal to \(-\left(\frac{n(n+1)}{2}\right)^{2}\). Reason: \(f(x)=\cos \frac{n(n-1)}{2} x+i \sin \frac{n(n-1)}{2} x\)

If for a non-zero \(x\), the function \(f(x)\) satisfies the equation \(a f(x)+b f\left(\frac{1}{x}\right)=\frac{1}{x}-5(a \neq b)\) then \(f^{\prime}(x)\) is equal to (A) \(\frac{1}{b^{2}-a^{2}}\left(\frac{a}{x^{2}}+b\right)\) (B) \(\frac{1}{a^{2}-b^{2}}\left(\frac{a}{x^{2}}+b\right)\) (C) \(\frac{1}{a^{2}-b^{2}}\left(\frac{a}{x^{2}}-b\right)\) (D) None of these

Suppose \(f(x)\) is differentiable \(x=1\) and \(\lim _{h \rightarrow 0} \frac{1}{h} f(1+h)=5\), then \(f^{\prime}(1)\) equals (A) 3 (B) 4 (C) 5 (D) 6

If \(f(x)=(1-x)^{n}\), then the value of \(f(0)+f^{\prime}(0)+\frac{f^{\prime \prime}(0)}{2 !}+\ldots+\frac{f^{n}(0)}{n !}\) is (A) \(n\) (B) 0 (C) \(2^{n}\) (D) \(2^{n}-1\)

If \(f\) is a real-valued differentiable function satisfying \(|f(x)-f(y)| \leq(x-y)^{2}, x, y \in R\) and \(f(0)=0\), then \(f(1)\) equals (A) \(-1\) (B) 0 (C) 2 (D) 1
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