f(x)=x2sin(x1)
f(0)=0;f(1)=sin(1)
In the interval (0,1):
0≤sin(x1)≤1
0≤x2sin(x1)≤x2
So,the function doesn't blow up to infinity,thus, f(x) is bounded.
(b)f(x)=xsinx
f(0)=0;f(1)=sin(1)
0≤sinx≤1
0≤xsinx≤x
So,the function doesn't blow up to infinity,thus, f(x) is bounded.
Comments
Dear Sheela John, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!
Thank you for your help assignment expert
Leave a comment