Finding the maximum of a continuous function $f:[1,7] \to \mathbb{R}$ is a continuous function. We know $f(1)=3$ and for every $x \in [1,7], f(x) \neq 10$.  The question is to prove the maximum of this function is less than $10$.
I honestly don't even know where to begin. 
 A: Suppose the maximum of $f$ were $\geq 10$. Since $f$ is continuous on a compact interval, there is a point $x_0 \in [1,7]$ such that
$$
f(x_0) = \sup\{f(x) : x\in [1,7]\}
$$
By hypothesis, $f(x_0) \neq 10$, and so $f(x_0) > 10$. But
$$
f(1) = 3 < 10 < f(x_0)
$$
Hence, by the intermediate value theorem, there is a point $y \in (1,x_0)$ such that $f(y) = 10$.
This is a contradiction, and hence $\sup\{f(x) : x\in [1,7]\} < 10$
A: IVT and extreme value theorem. Check that conditions are satisfied and then try to prove the result.
A: If a function is continuous then it must take on all values between any two points of its range. If it never hits 10 then it cannot have hit anything higher either because it would have had to pass through 10 to get there. 
You can also get this by contradiction: assume f never hits 10 but does hit 11 somewhere on its domain. Then there exists a point on the curve where $\lim\limits_{x\rightarrow x_0} f(x) \neq f(x_0)$ since we can find  a point on its domain where the function "skips" over 10 and possibly other values, to get to 11. However we assumed that f was continuous, so it must be false that it hits 11 without hitting 10. Therefore if it never hits 10 it cannot hit anything higher either. 
