Shiny_zubat wrote:

SomeonecoolLovesMaths wrote:

Define $g(x) = x - f(x)$ . Note that $g$ is differentiable.
$g'(x) = 1 - f'(x)$ $\forall x \in \mathbb{R}$ , so $\frac{3}{2} > g'(x) > \frac{1}{2}$ $\forall x \in \mathbb{R}$ .
Appying LMVT on the functions $g$ , $\forall x \in \mathbb{R}$ , $\exists c$ such that $g(x) - g(0) = g'(c) ( x - 0)$ Observe that, $\frac{x}{2} + g(0) < xg'(c) + g(0) = f(x) = xg'(c) + g(0) < \frac{3x}{2} + g(0)$ .
For $x = \frac{-2g(0)}{3} - 1$ , $g(x) < 0$ .
For $x = -2g(0) +1$ , $g(x) >0$ .
Applying IVT on $h$ , $\exists x_0 \in \mathbb{R}$ such that $g(x_0) = 0 = x_0 - f(x_0)$ .
Giving us $\boxed{f(x_0) = x_0}$ .

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Yes that was an error from my end @above, thanks for pointing that out! Corrected!