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Chapter 12: Problem 120

Let \(f: R \rightarrow R\) be a function defined by \(f(x)=\min\) \(\\{x+1,|x|+1\\} .\) Then which of the following is true? [2007] (A) \(f(x) \geq 1\) for all \(x \in \mathrm{R}\) (B) \(f(x)\) is not differentiable at \(x=1\) (C) \(f(x)\) is differentiable everywhere (D) \(f(x)\) is not differentiable at \(x=0\)

Short Answer

Expert verified
(A) is true, as \(f(x) \geq 1\) for all \(x\).

Step by step solution

01

Understanding the Function

The given function is defined as the minimum of two expressions: \(f(x) = \min\{x+1, |x|+1\}\). This means for any input \(x\), the output will be the smaller value between \(x+1\) and \(|x|+1\).
02

Analyzing the range of \(|x|+1\)

\(|x|\) is the absolute value of \(x\), so \(|x| + 1\) is always greater than or equal to 1. This is because the smallest value \(|x|\) can take is 0, making \(|x| + 1 = 1\).
03

Analyzing the range of \(x+1\)

For \(x+1\), it can be less than 1, equal to 1, or more than 1, depending on \(x\). Therefore, \(x+1\) can be less than \(|x|+1\) when \(x < 0\).
04

Determining Overall Range of \(f(x)\)

Since \(f(x) = \min\{x+1, |x|+1\}\), and \(|x|+1\) is always at least 1, this means the smallest \(f(x)\) can be is 1. Therefore, option (A), \(f(x) \geq 1\) for all \(x\), is true.
05

Checking Differentiability at Critical Points (0 and 1)

Critical points occur where the expressions inside the \(\min\) function intersect or where they are potentially undefined. We check differentiability at the points \(x = 0\) and \(x = 1\).
06

Evaluating at \(x = 0\)

At \(x = 0\), \(f(x) = \min\{0+1, |0|+1\} = 1\). As \(f(x)\) is constant around \(x = 0\), it is differentiable at this point.
07

Evaluating at \(x = 1\)

At \(x = 1\), we evaluate \(f(x) = \min\{1+1, |1|+1\} = 2\). However, detrivation of both expressions \(x+1\) and \(|x|+1\) near this point reveal that there is no sharp difference or discontinuity, but \(f(x) = x+1\) for \(x \geq 0\) around this region, so it is differentiable.
08

Drawing Conclusion

Upon analysis, since \(f(x)\) is differentiable at x=0 and x=1, option (D), stating that \(f(x)\) is not differentiable at \(x = 0\), is incorrect. Option (A) is correct as \(f(x) \geq 1\) for all \(x \in \mathbb{R}\).

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Key Concepts

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

Function Analysis
Function analysis involves understanding the behavior and characteristics of a function. When we approach the function \(f(x) = \min\{x+1, |x|+1\}\), we analyze the expression to determine outputs based on inputs. This function incorporates a minimum operator, making the output unique depending on the conditions applied to the input.
  • **Input Behavior:** Here, \(x+1\) is a straightforward linear expression, increasing with \(x\). \(|x|+1\), the absolute value expression, ensures results are non-negative. These behaviors impact the overall function output.
  • **Intersection Points:** Critical analysis includes identifying intersections or switch points where function behavior might change, crucial for investigating differentiability.
  • **Comparative Value:** Observing how the values of \(x+1\) and \(|x|+1\) compare decides which output the function adopts.
Understanding these aspects lets us predict and conclude characteristics such as continuity or differentiability effectively.
Minimum Function
A minimum function selects the smallest value from a set of function values, dictated by input conditions.
In our case \(f(x) = \min\{x+1, |x|+1\}\), the function chooses either \(x+1\) or \(|x|+1\) based on their comparison for the given \(x\).
  • **At \(x = 0\):** Both expressions are equal, so \(f(x) = 1\).
  • **When \(x < 0\):** Since \(x+1\) can become negative, and \(|x|+1\) remains positive, \(f(x)\) often selects \(x+1\) for negative \(x\) values.
  • **When \(x \geq 0\):** \(x+1\) is typically less than or equal to \(|x|+1\), so normally \(f(x)\) equals \(x+1\). This influences our understanding of function range and approach on both sides of critical points.
Grasping how and why the minimum is chosen provides insights into function behavior and key points that might affect differentiability.
Absolute Value Function
The absolute value function, \(|x|\), returns the non-negative value of \(x\). It plays a critical role in function analysis when combined with other terms, like in \(|x|+1\). This expression behaves differently from a purely linear function, and understanding this is key for analysis.
  • **Non-negative Output:** \(|x|\) ensures the function is non-negative, inherently pushing \(|x|+1\) to be greater than or equal to 1.
  • **Break Points:** Critical when defining a minimum or conditions for differentiability because there are distinct cases for \(x < 0\) and \(x \geq 0\).
  • **Combining Characteristics:** in this exercise, \(|x|+1\) is constantly at odds with \(x+1\), adding complexity to deciding function characteristics such as slope and symmetry.
Recognizing how absolute values interact with linear terms within a minimum function allows us to effectively address potential differentiability issues, notably around zero derivatives and piecewise distribution of values.

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Most popular questions from this chapter

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: Let \(f(x+y)=f(x) f(y)\) for all \(x, y\), where \(f(0) \neq 0 .\) If \(f^{\prime}(0)=2\), then \(f(x)=A e^{2 x}\), where \(A\) is a constant. Reason: \(f^{\prime}(x)=f(x)\)

Let \(f\) be a continuous function on \(R\) such that \(f(1 / 4 n)=\left(\sin e^{n}\right) e^{-n^{2}}+\frac{n^{2}}{n^{2}+1} .\) Then, the value of \(f(0)\) is (A) 1 (B) \(\frac{1}{2}\) (C) 0 (D) None of these

Let \(g(x)=x f(x)\), where \(f(x)=\left\\{\begin{array}{ll}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.\). At \(x=0\), (A) \(g\) is differentiable but \(g^{\prime}\) is not continuous (B) \(\mathrm{g}\) is differentiable while \(f\) is not (C) both \(f\) and \(g\) are differentiable (D) \(g\) is differentiable and \(g^{\prime}\) is continuous

Let \(f: R^{+} \rightarrow R\) satisfies the equation \(f(x y)=e^{x y-x-y}\) \(\left\\{e^{y} f(x)+e^{x} f(y)\right\\}, \forall x, y \in R^{+} .\)If \(f^{\prime}(1)=e\), then \(f(x)=\) (A) \(e^{x} \log |x|\) (B) \(e^{-x} \log |x|\) (C) \(e^{2 x} \log |x|\) (D) None of these

Let \(f: R \rightarrow R\) be a function such that \(f\left(\frac{x+y}{3}\right)=\frac{f(x)+f(y)}{3}, f(0)=0\) and \(f^{\prime}(0)=3\) Then, (A) \(f(x)\) is a quadratic function (B) \(f(x)\) is continuous but not differentiable (C) \(f(x)\) is differentiable in \(R\) (D) \(f(x)\) is bounded in \(R\)
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