Let $ f(x) = xe^{-x} $. I need to prove that $ f([0, 1]) = [0, \frac{1}{e}] $. Can you verify my proof?
$ \underline{[0, \frac{1}{e}] \subseteq f([0, 1]):} $
First, $ f(0) = 0e^0 = 0, f(1) = 1\cdot e^{-1}=\frac{1}{e}$. Let $ x \in (0, \frac{1}{e}) $. Therefore, $ f(0) \le x \le f(1) $. $ \ f \ $ is a continuous function in the closed interval $ [0, 1] $ as the composition and sum of continuous functions. By IVT, we get that $ \exists c \in [0, 1](f(c) = x) $. Hence, $ [0, \frac{1}{e}] \subseteq f([0, 1]) $.
$ \underline{f([0,1]) \subseteq [0, \frac{1}{e}]:} $
Let $ x \in [0, 1] $. If $ x = 0 $, then $ f(x) = 0 \in [0, \frac{1}{e}] $. If $ x = 1 $, then $ f(x) = \frac{1}{e} \in [0, \frac{1}{e}]. \ $ $ x > 0 \land e^{-x} > 0 \Rightarrow xe^{-x} \Rightarrow f(x) > 0 $.
Suppose, to the contrary, that $ f(x) \ge \frac{1}{e} $. We will show that there exists a $ b \in (0, 1) $ such that $ f(b) = \frac{1}{e} $. If $ f(x) = \frac{1}{e} $, choose $ b = x $.If $ f(x) \gt \frac{1}{e} $, then, by the IVT, there exists $ c \in [0, x] $ such that $ f(c) = \frac{1}{e} $, choose $ b = c $.
$ f $ is differentiable in the open interval $ (b, 1) $ as the composition and sum of differentiable functions, and it is also continuous on $ [b, 1] $. $ f(b) = f(1) $, which means there exists $ d \in (b, 1) $ such that $ f^{'}(d) = 0 $ by Rolle's Theorem.
We will check when $ f^{'}(x) = 0 $.
$ f^{'}(x) = e^{-x} - xe^{-x} = 0 $
$ e^{-x}(1 - x) = 0 \Rightarrow x = 1 $.
But $ d \neq 1 $ - contradiction.
Therefore, $ 0 < f(x) < 1 \Rightarrow \forall x_0 \in f([0,1]), x_0 \in [0, \frac{1}{e}] $.
$$f’(x) = (1-x)e^{-x}$$ For $0 \le x<1$, $f’(x) > 0$ which implies that function is strictly increasing for the interval $[0,1)$ and clearly $f’(1)=0$ and $f(x)$ is continuous for all real $x$ so in the interval [0,1], $${f(x)}_{min} = f(0) = 0$$ $${f(x)}_{max}= f(1)=\frac{1}{e}$$