TITLE: $f'+\lambda f$ acts as $f'$ without integral
QUESTION [7 upvotes]: Problem
Suppose that $f(x)$ is a differentiable function and $f'(x)$ is the derivative of $f(x)$. Without aids of integral, can we prove that $f'(x)+\lambda f(x)$ has intermediate property?
Intermediate property
A (real) function $f(x)$ having intermediate property means that if $a,b\in f(\Bbb R)$ and $a<c<b$, then $c\in f(\Bbb R)$, where $f(\Bbb R)=\{\;f(x):x\in\Bbb R\;\}$
With aids of integral
Let $g(x)=f'(x)+\lambda f(x)$. For $f(x)$ is continuous, we have $f(x)$ is Riemann-integrable. Let $G(x)=f(x)+\int_0^xf(t)dt$, we have $G'(x)=g(x)$; therefore, we can apply Darboux's theorem to $G(x)$, and we've done.

REPLY [3 votes]: Let $a\lt b$, $g(t)=f'(t)+\lambda f(t)$, and $z$ in the interval between $g(a)$ and $g(b)$. One wants to show that there exists $x$ in $[a,b]$ such that $g(x)=z$. If $\lambda=0$, this is the usual case. Assume that $\lambda\ne0$.
One can assume without loss of generality that $g(a)\lt z\lt g(b)$.  Consider the function $h$ defined by  $h(t)=\mathrm e^{\lambda t}(f(t)-z/\lambda)$. Then $h'(t)= \mathrm e^{\lambda t}(g(t)-z)$, hence $h'(a)\lt0\lt h'(b)$. The usual intermediate value theorem for derivatives applied to $h'$ ensures that $h'(x)=0$ for some $x$ in $(a,b)$. Thus, $g(x)=z$.