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!

  • $\begingroup$ Hint: $0<kc<1$ is always possible $\endgroup$ – imranfat Nov 25 '16 at 6:03

Consider the function $$g(x)=f(x)-kx$$ $g(x)$ is continuous, and satisfies $g(0)=1>0$ and $g(1)=-k<0$. Now apply IVT to get some $c$ in $(0,1)$ with $g(c)=0$.

Note: $f(x)$ need not be differentiable.

  • $\begingroup$ I don't see why $g(0)=1$and why $k<0$ ... $\endgroup$ – Michael Hoppe Nov 25 '16 at 6:06
  • $\begingroup$ Because $f(0)=1$ (given) and $k*(0)=0$ $\endgroup$ – imranfat Nov 25 '16 at 6:08
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    $\begingroup$ Omg, it's too early here in Germany to put comments ... I read $f(0)=0$. $\endgroup$ – Michael Hoppe Nov 25 '16 at 6:10

Consider $g(x)=f(x)-kx$. Note that $g(0)=1>0$ and $g(1)=-k<0$. 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 $f(c)=kc$


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