Calculus -2

Taking the integrating factor $p(x) = e^{x}$ , multiplying the original inequality through by this factor results in:

$p(x)[f(x) + f'(x) \le 1] \Rightarrow e^{x}f(x) + e^{x}f'(x) \le e^{x}$ ;

or $[e^{x}f(x)]' \le e^{x};$

or $e^{x}f(x) \le e^{x} + C$ ;

or $f(x) \le Ce^{-x} + 1$

Given that $f(0) = 0$ , we solve for the real constant C: $0 \le Ce^{0} + 1 \Rightarrow -1 \le C$ , which produces $f(x) \le 1 - e^{-x}$ . The maximum value of $f(1)$ computes to $f(1) \le 1 - \frac{1}{e} \approx \boxed{0.63212}.$