Question

Show that `f` is strictly increasing in `RR` ?

`f:RR->RR` differentiable with `f'` continuous in `RR`, `f(0)=0` , `f(1)=1`

• `f(f(x))+f(x)=2x` ,

`AA` `x` `in` `RR`

Show that `f` is strictly increasing in `RR`

Answer 1

Sign/contradiction & Monotony

Explanation:

`f` is differentiable in `RR` and the property is true `AAx` `in` `RR` so by differentiating both parts in the given property we get

`f'(f(x))f'(x)+f'(x)=2` (1)

If `EEx_0` `in` `RR:f'(x_0)=0` then for `x=x_0` in (1) we get

`f'(f(x_0))cancel(f'(x_0))^0+cancel(f'(x_0))^0=2` `<=>`

`0=2` `->` Impossible

Hence, `f'(x)!=0` `AA` `x` `in` `RR`

• `f'` is continuous in `RR`
• `f'(x)!=0` `AA` `x` `in` `RR`

`->` `{(f'(x)>0", "),(f'(x)<0", "):}` `x` `in` `RR`

If `f'(x)<0` then `f` would be strictly decreasing

But we have `0<1` `<=>^(fdarr)` `<=>` `f(0)>f(1)` `<=>`

`0>1` `->` Impossible

Therefore, `f'(x)>0`, `AA` `x` `in` `RR` so `f` is strictly increasing in `RR`