$f(0) = 1$ and $f(1) = 0$ Show there exists $c$ in $(0, 1)$ such that $f(c) = kc$. Let $f$ be continuous on $[0, 1]$ and differentiable on $(0, 1)$ such that $f(0) = 1$ and $f(1) = 0$ and
let $k$ be a positive constant.

Show that there exists $c$ in $(0, 1)$ such that $f(c) = kc$.

I understand I need to substitute $f$ by another function $g$ and apply Intermediate Value Theorem, but I am clueless how to transform $f$. Any kind soul please help!
Consider $g(x)=f(x)-kx$. Note that $g(0)=1&gt;0$ and $g(1)=-k&lt;0$. Then apply Intermediate Value Theorem.
Define $g(x)=f(x)-kx$ . Then, $g(0)=1$ and $g(1)=-k$. Then apply Intermediate Value Property for $g$ there exists $c \in (0,1)$ such that $g(c)=0$. That is $f(c)=kc$