Suppose f: IR-IR is a moNotonically increas ing function.(a) Prave if tie fix-fix-1) = 0 , then fix) = 0Lim flx+t) - flx-h)0 , then f'xo = 0of(6) Give an example of a function 9 and a point x€ IR such that%3D%3DLim 9(x+h) -g (x-h)h-0=0 but g'(x) does not exist

Name: Suppose f: IR-IR is a moNotonically increas ing function.(a) Prave if
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Description: Suppose f: IR-IR is a moNotonically increas ing function.(a) Prave if tie fix-fix-1) = 0 , then fix) = 0Lim flx+t) - flx-h)0 , then f'xo = 0of(6) Give an example of a function 9 and a point x€ IR such that%3D%3DLim 9(x+h) -g (x-h)h-0=0 but g'(x) doe

Answer for sub question a: Suppose f:ℝ→ℝ is a monotonically increasing function and suppose…

Answered: Suppose f: IR→IR is a moNotonically…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Suppose f: IR-IR is a moNotonically increas ing function.
(a) Prave if tie fix-fix-1) = 0 , then fix) = 0
Lim flx+t) - flx-h)
0 , then f'xo = 0
of
(6) Give an example of a function 9 and a point x€ IR such that
%3D
%3D
Lim 9(x+h) -g (x-h)
h-0
=0 but g'(x) does not exist

Expert Solution

Step 1

Answer for sub question a:

Suppose $f : ℝ \to ℝ$ is a monotonically increasing function and suppose $\lim_{h \to 0} \frac{f (x + h) - f (x - h)}{2 h} = 0$

Then since $f$ is monotonically increasing it is differentiable almost everywhere and since $\lim_{h \to 0} \frac{f (x + h) - f (x - h)}{2 h} = 0$ it follows that $f$ is differentiable at $x$ .

Now, since $f$ is differentiable at $x$ ,

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Now,

$\begin{array}{rcl} \frac{f (x + h) - f (x - h)}{2 h} & = & \frac{f (x + h) - f (x) + f (x) - f (x - h)}{2 h} \\ = & \frac{1}{2} (\frac{f (x + h) - f (x)}{h} + \frac{f (x) - f (x - h)}{h}) \end{array}$

Hence,

$\begin{array}{rcl} \lim_{h \to 0} \frac{f (x + h) - f (x - h)}{2 h} & = & \frac{1}{2} (\lim_{h \to 0} \frac{f (x + h) - f (x)}{h} + \lim_{h \to 0} \frac{f (x) - f (x - h)}{h}) \\ = & \frac{1}{2} (f' (x) + f' (x)) \\ = & f' (x) \end{array}$

Therefore,

$\begin{array}{rcl} f' (x) & = & \lim_{h \to 0} \frac{f (x + h) - f (x - h)}{2 h} \\ = & 0 \end{array}$

Hence the claim.

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