Question

suppose f is function with property
\[\large |f(x)|\leq x^2\] for any x in R

a) show that f(0)=0
b) show that f'(0)=0

Answer 1

so far no idea hehe :)

Answer 2

Here's my guess, but I'm not entirely sure it's allowed since I think the square has to be under the square root, not inside out like this: \[\Large \sqrt{f(x)}^2 \le x^2 \\ \Large \pm \sqrt{f(x)}\le x\]

Answer 3

|f(x)| <= x^2

turns into

-x^2 <= f(x) <= x^2

then you plug in x = 0

-x^2 <= f(x) <= x^2

-0^2 <= f(0) <= 0^2

0 <= f(0) <= 0

so f(0) = 0

I'm not sure how to do part b

Answer 4

sounds good to me @Kainui

Answer 5

differentiate !

Answer 6

your inequality

Answer 7

-2x<f'(x)<2x

Answer 8

damn @jim_thompson5910 was thinking to do that thought it wouldn't work
awesome!!

Answer 9

is that allowable? @freckles

Answer 10

freckles, you forgot about the absolute value

Answer 11

well that inequality is true for x>0

and then 2x<f'(x)<-2x for x<0

Answer 12

if y = |x|, then dy/dx = |x|/x or dy/dx = x/|x|

Answer 13

\[-x^2 \le f(x) \le x^2 \]
so we can't just differentiate everything here?

Answer 14

hmm doing cases
yes that good! both way we get f'(0)=0

Answer 15

oh true, just break it up and then differentiate, nvm then

Answer 16

yeah makes sense to me, awesome guys ^^