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

Question

$f:RR->RR$ 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 $RR$

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Answer 1 · 2018-05-24T18:37:34

Sign/contradiction & Monotony

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

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

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 $RR$