/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 It is easy to find a function \(... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 62

It is easy to find a function \(f\) such that \(|f|\) is differentiable but \(f\) is not. For example, we can choose \(f(x)=1\) for \(x\) rational and \(f(x)=-1\) for \(x\) irrational. In this example \(f\) is not even continuous, nor is this a mere coincidence: Prove that if \(|f|\) is differentiable at \(a,\) and \(f\) is continuous at \(a\) then \(f\) is also differentiable at \(a\). Hint: It suffices to consider only \(a\) with \(f(a)=0 .\) Why? In this case, what must \(|f|^{\prime}(a)\) be?

Short Answer

Expert verified
If \(|f|\) is differentiable and \(f\) continuous at \(a,\) then \(f\) is also differentiable at \(a\) with \(f^{\prime}(a) = 0\).

Step by step solution

01

Understand the problem statement

We need to show that if \(|f|\) is differentiable at \(a\) and \(f\) is continuous at \(a\), then \(f\) must also be differentiable at \(a\). Consider the special case where \(f(a) = 0\).
02

Express differentiability

Differentiability of \(|f|\) at \(a\) means the limit as \(h \to 0\) of \(\frac{|f(a+h)| - |f(a)|}{h}\) exists. Since \(f(a) = 0\), this simplifies to \(\frac{|f(a+h)|}{h}\).
03

Use the continuity of \(f\)

Since \(f\) is continuous at \(a\), for every \(\forall \epsilon > 0\), there exists a \(\forall \delta > 0\) such that if \(0 < |h| < \delta\), then \(|f(a+h) - f(a)| = |f(a+h)| < \epsilon\).
04

Examine the derivative of \(|f|\)

Given \(f(a) = 0\), we have \(\frac{|f(a+h)|}{h}\) should approach 0 as \(h \to 0\) if \(|f|\) is differentiable at \(a\). Therefore, \(|f|^{\prime}(a) = 0\).
05

Show the difference quotient for \(f\)

Since \(|f|\) is differentiable at \(a\) and \(f(a) = 0\), we have \(\frac{f(a+h) - f(a)}{h} = \frac{f(a+h)}{h}\) approaching 0. Thus, \(f\) is differentiable at \(a\) with \(f^{\prime}(a) = 0\).
06

Conclude the proof

Since the limit defining the derivative of \(f\) exists and equals 0, we conclude that if \(|f|\) is differentiable at \(a\) and \(f\) is continuous at \(a\), then \(|f|\) must also be differentiable at \(a\) with \(f^{\prime}(a) = 0\).

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Key Concepts

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

absolute value function
An absolute value function is written as \(|f(x)|\). It transforms every input into its non-negative counterpart. For example, if \(f(x) = x\), then \(|f(x)| = |x|\). The absolute value function is piecewise-defined:
  • \(f(x) \ge 0\): output is \(|x| = x\)
  • \(f(x) < 0\): output is the negation of \(|x| = -x\)
The key property is that \(|x| \) keeps outputs non-negative, no matter the input. If given \( f(x)\), the graph of \(|f(x)|\) will either be the same or a reflection above the x-axis.
continuity
Continuity of a function at a point means there are no jumps, gaps, or breaks at that point. Mathematically, \(f\) is continuous at \(a\) if:
  • The function is defined at \(a\): \(f(a)\) exists
  • The limit as \(x \to a\) of \(f(x)\) exists
  • The value of the limit is equal to the function's value at that point: \(\lim_{{x \to a}} f(x) = f(a)\)
In our problem, we know \(f\) is continuous at \((a)\), which will help us connect the differentiability between \(|f|\) and \(f\).
derivative limits
The core of differentiability lies in the limit definition of a derivative. A function \(f\) is differentiable at \(a\) if:
\[\lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h} \text{ exists}\]When considering an absolute value function \(|f|\), we look at the limit:
  • If \(f(a) = 0\), then the derivative limit test becomes \(\frac{|f(a+h)|}{h}\)
  • The existence of this limit and its approach to 0 suggests the slope is smooth at point \((a)\)
This understanding solidifies how we transition from \(|f|\) to \(f\).
step-by-step proof
Let's walk through the proof:
  • Step 1: Set up the problem by acknowledging we're proving \(\textrm{If } |f| \textrm{ is differentiable and } f \textrm{ is continuous at } a, \textrm{ then } f \textrm{ is also differentiable.}\)
  • Step 2: Define differentiability of \(|f|\) using the limit \(\lim_{{h \to 0}} \frac{{|f(a+h)|}}{h}\)
  • Step 3: Utilize continuity to connect the behavior of \(f\) at point \((a)\) with \(\lim_{{h \to 0}} \frac{{|f(a+h)|}}{ h }\)
  • Step 4: Simplify to confirm \(\lim_{{h \to 0}} \frac{{|f(a+h)|}}{ h } = 0\)
  • Step 5: Prove \(f\) is differentiable using similar derivatives limits, leading us to deduce \(\lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{ h } = \lim_{{h \to 0}} \frac{{f(a+h)}}{ h } = 0\)
  • Conclusion: The differentiability of \(|f|\) and continuity of \(f\) show that \(|f|\) and \(f\) have the same derivatives properties.
calculus
Calculus is the branch of mathematics that deals with limits, functions, derivatives, integrals, and infinite series. Our problem engages calculus concepts such as:
  • Differentiability: It shows through limits and derivative definitions.
  • Continuity: Ensures there's no break in the function's graph at point \(a\).
  • The Absolute Value Function: Transforms function values into non-negatives, keeping graphs above the x-axis.
Understanding these calculus principles helps us follow proofs for higher-level math problems involving differentiability and continuity.

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Most popular questions from this chapter

Find the trapezoid of largest area that can be inscribed in a semicircle of radius \(a,\) with one base lying along the diameter.

Find the following limits: (i) \(\lim _{x \rightarrow 0} \frac{x}{\tan x}.\) (ii) \(\lim _{x \rightarrow 0} \frac{\cos ^{2} x-1}{x^{2}}.\)

(a) Suppose that the polynomial function \(f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\) has exactly \(k\) critical points and \(f^{\prime \prime}(x) \neq 0\) for all critical points \(x\). Show that \(n-k\) is odd. (b) For each \(n,\) show that if \(n-k\) is odd, then there is a polynomial function \(f\) of degree \(n\) with \(k\) critical points, at each of which \(f^{\prime \prime}\) is non-zero. (c) Suppose that the polynomial function \(f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{0}\) has \(k_{1}\) local maximum points and \(k_{2}\) local minimum points. Show that \(k_{2}=k_{1}+1\) if \(n\) is even, and \(k_{2}=k_{1}\) if \(n\) is odd. (d) Let \(n, k_{1}, k_{2}\) be three integers with \(k_{2}=k_{1}+1\) if \(n\) is even, and \(k_{2}=k_{1}\) if \(n\) is odd, and \(k_{1}+k_{2} < n .\) Show that there is a polynomial function \(f\) of degree \(n,\) with \(k_{1}\) local maximum points and \(k_{2}\) local minimum points. Hint: Pick \(a_{1} < a_{2} < \cdots< a_{k_{1}+k_{2}}\) and try \(f^{\prime}(x)=\prod_{i=1}^{k_{1}+k_{2}}\left(x-a_{i}\right) \cdot\left(1+x^{2}\right)^{l}\) for an appropriate number \(l.\)

A function \(f\) is increasing at \(a\) if there is some number \(\delta>0\) such $$f(x)>f(a) \text { if } a< x< a+\delta$$ and $$f(x)< f(a) \text { if } a-\delta < x < a$$ Notice that this does not mean that \(f\) is increasing in the interval \((a-\delta .\) \(a+\delta) ;\) for example, the function shown in Figure 34 is increasing at \(0,\) but is not an increasing function in any open interval containing 0. (a) Suppose that \(f\) is continuous on [0.1] and that \(f\) is increasing at \(a\) for every \(a\) in \([0.1] .\) Prove that \(f\) is increasing on \([0.1] .\) (First convince yourself that there is something to be proved.) Hint: For \(0< b <1\) prove that the minimum of \(f\) on \([b, 1]\) must be at \(b\). (b) Prove part (a) without the assumption that \(f\) is continuous, by considering for each \(b\) in [0.1] the set \(S_{b}=\\{x: f(y) \geq f(b) \text { for all } y \text { in }[b, x]\\}\) (This part of the problem is not necessary for the other parts.) Hint: Prove that \(S_{b}=\\{x: b \leq x \leq 1\\}\) by considering sup \(S_{b}.\) (c) If \(f\) is increasing at \(a\) and \(f\) is differentiable at \(a\), prove that \(f^{\prime}(a) \geq 0\) (this is easy). (d) II \(f^{\prime}(a)>0,\) prove that \(f\) is increasing at \(a\) (go right back to the definition of \(f^{\prime}(a)\).) (e) Use parts (a) and (d) to show, without using the Nean Value Theorem, that if \(f\) is continuous on \([0,1]\) and \(f^{\prime}(a)>0\) for all \(a\) in \([0.1],\) then \(f\) is increasing on \([0,1].\) (f) Suppose that \(f\) is continuous on [0,1] and \(f^{\prime}(a)=0\) for all \(a\) in (0,1) Apply part (e) to the function \(g(x)=f(x)+\varepsilon x\) to show that \(f(1)-\) \(f(0)>-\varepsilon .\) Similarly, show that \(f(1)-f(0)<\varepsilon\) by considering \(h(x)=\varepsilon x-f(x) .\) Conclude that \(f(0)=f(1).\) This particular proof that a function with zero derivative must be constant has many points in common with a proof of H. A. Schwarz, which may be the first rigorous proof ever given. Its discoverer, at least, seemed to think it was. See his exubcrant letter in reference [54] of the Suggested Reading.

(a) Prove that the function \(f(x)=x^{2}-\cos x\) satisfies \(f(x)=0\) for precisely two numbers \(x.\) (b) Prove the same for the function \(f(x)=x^{2}-x \sin x-\cos x.\) (c) Prove this also for the function \(f(x)=2 x^{2}-x \sin x-\cos ^{2} x .\) (Some preliminary estinates will be useful to restrict the possible location of the zeros of \(f .)\)
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