Answer to Question #248042 in Calculus for Asma

Question #248042

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
2021-10-08T04:40:28-0400

(a)f(x)=x2sin(1x)(a) 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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