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Suppose $f: [1,2] \to [5,7]$ is continuous. Show that $f(c)=2c+3$ for some $c \in [1,2]$.

First note $f(1)=5$ and $f(2)=7$. By the IVT, all values $c \in [1,2]$ are hit. I'm just wondering how to put all of these facts together to arrive at $f(c)=2c+3$ for all $c \in [1,2]$.

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You can't prove that $f(c) = 2c+3$ for all $c \in [1,2]$, merely that there exists such $c$. E.g. $f(x) = 2x+3+\sin(2\pi x)$ also fulfills the premises. – dtldarek Dec 3 '12 at 8:20

Look at the function $$g(x) = f(x) - 2x - 3$$ We have $g(1) = f(1) - 5 \geq 0$ and $g(2) = f(2) - 7 \leq 0$ since $f(x) \in [5,7]$ for all $x \in [1,2]$.

Hence, by intermediate value theorem, there is a $c \in [1,2]$ such that $g(c) = 0$. Hence, there exists a $c \in [1,2]$ such that $$f(c) = 2c+3$$

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