Answer to Question #114832 in Real Analysis for Sheela John

Question #114832
Determine which of the following functions are of bounded variation on[0,1]
a. f(x)= x^2sin(1/X) if x#0 ,f(0)=0
b. f(x)= √x sinx,if x#0,f(0)=0
1
Expert's answer
2020-05-11T08:51:23-0400

f(x)=x2sin(1x)f(x)=x^2sin(\frac1x)

f(0)=0;f(1)=sin(1)f(0)=0;f(1)=sin(1)

In the interval (0,1):

0sin(1x)10\leq sin(\frac1x)\leq 1

0x2sin(1x)x20\leq x^2sin(\frac1x)\leq x^2

So,the function doesn't blow up to infinity,thus, f(x) is bounded.

(b)f(x)=xsinxf(x)=\sqrt{x}sin x

f(0)=0;f(1)=sin(1)f(0)=0;f(1)=sin(1)

0sinx10\leq sinx \leq1

0xsinxx0\leq \sqrt{x}sinx \leq\sqrt x

So,the function doesn't blow up to infinity,thus, f(x) is bounded.


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Comments

Assignment Expert
12.05.20, 20:50

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Sheela John
12.05.20, 10:55

Thank you for your help assignment expert

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