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Magazine Chapter 12: Problem 27
Let \(f(x)=a[x]+b e^{|x|}+c|x|^{2}\), where \(a, b\) and \(c\) are real constants. If \(f(x)\) is differentiable at \(x=0\), then (A) \(b=0, c=0, a \in R\) (B) \(a=0, c=0, b \in R\) (C) \(a=0, b=0, c \in R\) (D) None of these
Short Answer
Expert verified
The correct answer is (C): \(a=0, b=0, c \in R\).
Step by step solution
01
Understanding the Problem
We have the function \( f(x) = a[x] + b e^{|x|} + c|x|^{2} \). We need to determine the values of \(a\), \(b\), and \(c\) such that \(f(x)\) is differentiable at \(x = 0\).
02
Differentiability at a Point
A function is differentiable at a point if the derivative exists at that point. To find the derivative at \(x = 0\), the limit \( \lim_{h \to 0} \frac{f(h) - f(0)}{h} \) must exist.
03
Evaluating \( f(0) \)
At \(x = 0\), we need to evaluate \(f(0)\). Since \([0] = 0\) and \(|0| = 0\), we have \(f(0) = a[0] + b e^{|0|} + c|0|^2 = a(0) + b(1) + c(0) = b\). Thus, \( f(0) = b \).
04
Expression for \( f(h) \)
For small \(h\), \(f(h) = a[h] + b e^{|h|} + c |h|^2\). For small non-zero \(h\), \([h] = 0\) if \(h \geq 0\) and \([-1]\) if \(h < 0\).
05
Finding \( f(h) - f(0) \)
If \(h > 0\), \(f(h) - f(0) = a[0] + b e^{h} + c h^2 - b = b(e^h - 1) + ch^2\). If \(h < 0\), \(f(h) - f(0) = a(-1) + b e^{-h} + c h^2 - b = -a + b(e^{-h} - 1) + ch^2\).
06
Calculating the Limit for \(h > 0\)
Evaluate the limit \( \lim_{h \to 0^+} \frac{b(e^h - 1) + ch^2}{h} = b \lim_{h \to 0^+} \frac{e^h - 1}{h} + c \lim_{h \to 0^+} h \). We know \( \lim_{h \to 0^+} \frac{e^h - 1}{h} = 1 \), thus the limit is \(b + 0 = b\).
07
Calculating the Limit for \(h < 0\)
Evaluate the limit \( \lim_{h \to 0^-} \frac{-a + b(e^{-h} - 1) + ch^2}{h} = \lim_{h \to 0^-} \frac{-a}{h} + b \lim_{h \to 0^-} \frac{e^{-h} - 1}{h} + c \lim_{h \to 0^-} h \). The term \(-a/h\) leads to discontinuity unless \(a = 0\). Thus \(a = 0\); \( \lim_{h \to 0^-} \frac{e^{-h} - 1}{h} = -1 \), so it's \(-b\).
08
Ensuring Continuity
For the derivative to exist and be equal, both limits (from Step 6 and Step 7) must be the same, i.e., \(b = -b\). Thus, \(b = 0\).
09
Determining the Conditions for Differentiability
With \(a = 0\) and \(b = 0\), only the term \(c|x|^2\) affects differentiability, which is always differentiable regardless of \(c\). Thus, \(c\) can be any real number.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Functions
A piecewise function is a function defined by multiple sub-functions, each corresponding to a specific interval of the input variable. These functions can often appear complex due to their segmented nature but are essential for modeling situations where a rule changes based on input conditions. In the original exercise, the function is
- Defined with a function of various components like \[f(x) = a[x] + b e^{|x|} + c|x|^2\]
- Each component operates within a "piece" of the input domain, like \([x]\) which could behave differently depending on whether the input is negative or positive.
Continuity
Continuity is a fundamental property of functions that ensures there are no breaks, jumps, or holes in their graph. A function is said to be continuous at a point if the limit of the function as the input approaches the point equals the function's value at that point. For a function to be continuous:
- The function must be defined at the point.
- The limit of the function as it approaches the point from both sides must exist.
- The value of the limit must equal the function's value at that point.
Limit Calculations
Limit calculations are key to determining the behavior of a function as inputs get arbitrarily close to a certain point. Limits help in pinpointing how a function behaves near points of interest, which is crucial for both continuity and differentiability. To calculate the limit, consider:
- Approaching the point from various directions or sides, typically left and right.
- Determining whether the function tends towards a finite value.
- Using algebraic manipulations鈥攐r occasionally L'H么pital's rule鈥攖o evaluate difficult expressions.
Derivative at a Point
The derivative of a function at a point provides its instantaneous rate of change at that point. For a function to have a derivative at a point means that its graph has a tangent line without abrupt changes in direction at that point. Mathematically, this involves finding \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)and ensuring this limit exists. For the function given in the exercise, the derivative at \(x = 0\) focuses on eliminating terms that could prevent the existence of a derivative, such as terms leading to discontinuity or undefined limits. By applying the respective conditions to eliminate these terms, we ensure a smooth and calculable rate of change at the target point.
