Suppose that $f$ is a continuous function and that $f(0)=1$ and $f(1)=2$. Show that there is $c\in(0,1)$ such that $f(c)=-\ln (2-2c)$.
It is a bit different from the technique I used Here.
Any ideas?
A question on Intermediate Value Theorem
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calculus
2 Answers
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For continuous function $f(x)$, let $$g(x)=\dfrac12e^{-f(x)}+x-1$$ then $g$ is a continuous function and $g(0)=\dfrac{1}{2e}-1<0$ and $g(1)=\dfrac{1}{2e^2}>0$, so by IVT we get $\exists c\in(0,1)$ such that $\dfrac12e^{-f(c)}+c-1$ or $f(c)=-\ln(2-2c)$.
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Let $h(x) = f(x) + \ln(2-2x)$. $h$ is defined and continuous on $[0,1)$. We have $\lim_{x \to 1} h(x) = -\infty$, so there exists some $t \in (0,1)$ such that $h(t) < 0$. Now $h(0) = 1 +\ln2 > 0$, so by IVT, there is $c \in (0,1)$ such that $h(c)=0$.