Question

If |f"(x)|<=1 and f(0)=f'(0)=0
Then
find f(2) lie in the interval.......

Answer 1

Should I have to integrate first..

Answer 2

For some reason I'm thinking mean value theorem can be applied here

Answer 3

pls explain

Answer 4

This is as far as I'm able to get with that idea
\[f'(c)=\int\limits_0^c f''(t) dt=\frac{f(2)}{2} \\ \text{ where } c \in (0,2)\]
I don't know where to go from there
and if this is the right direction

Answer 5

I have assumed things that the question did not give like continuity and differentiableness on interval (0,2)

Answer 6

Can't we simply integrate it from there..
-1<f"(x)<1

Answer 7

actually that might work

Answer 8

By doing like this ...
I got
-x2/2<f(x)< x2/2

Answer 9

\[-x+C \le f'(x)+K \le x+D \\ \text{ subtract } K \text{ on all sides } \\ -x+(C-K) \le f'(x) \le x+(D-K) \\ -x+c \le f'(x) \le x+d \\ \text{ plug in } x=0 \\ \text {we get } -0+c \le f'(0) \le 0+d \\ \implies c=d=0 \text{ since } f'(0)=0 \\ \text{ so you have that } \\ -x \le f'(x) \le x \\ \text{ and you can do something similar for } f(x) \\ \text{ giving you that inequality you have there }\]

So plug in 2 for x now

Answer 10

Thanks ...I got it