I really would appreciate some help:
If f(x)=[(x^2)sin(1/x), x not = 0,
c, x=0]
c is a constant,
Fill in the blanks with polynomials:
_____≤(x^2)sin(1/x)≤_____

I have no idea how to even begin, please help

Question

I really would appreciate some help:
If f(x)=[(x^2)sin(1/x), x not = 0, 
             c, x=0]
c is a constant, 
Fill in the blanks with polynomials:
_____≤(x^2)sin(1/x)≤_____

I have no idea how to even begin, please help

jamesj · Answer

$$ | \sin x | \leq 1 $$ hence $$ | \sin(1/x) | \leq 1 $$

anonymous · Answer

would you mind explaining that a little more, sorry about the trouble

jamesj · Answer

sin(x) oscillates between -1 and 1.  So we can bound it below by -1 and above by +1

In fact, for any function f, sin(f(x)) still only varies between -1 and 1.

Given that, you should be able to see now how to bound your expression above and below

jamesj · Answer

To be explicit, for all $ x \neq 0 $,
$$ -1 \leq \sin(1/x) \leq 1 $$

jamesj · Answer

Here's the plot

anonymous · Answer

I see but the problem asks for polynomials I think that is what is confusing me, how would you find them?

jamesj · Answer

$ x^2 $ is a polynomial

anonymous · Answer

then the blanks would be -x^2 and x^2?

jamesj · Answer

yes

anonymous · Answer

I think i have a better idea now, thank you for your help and patience!

jamesj · Answer

here's another example,
$$ -x^2 \leq \sin(e^{-x^2} + x^{100}) x^2 \leq x^2 $$
This is true because sin can't be any more than +1 or any less than -1, despite whatever the form of the argument of sin(.) is.

anonymous · Answer

I understand, its because of the sin then, once again thank you!