So I need to prove that $$\frac{1}{16} <\sqrt{51} -7< \frac{1}{ 14}$$ using the MVT and no calculator but I can't find a function that would satisfy those outputs ![as for the question of where is the 2 clearly there isn't one that's exactly why I'm stuck]
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$\begingroup$The facts that $7 = \sqrt{49}$ and $49$ is close to $51$ should suggest using $f(x) = \sqrt{x}$. There exists $c \in (49,51)$ satisfying $$\frac{\sqrt{51} - 7}{2} = \frac{f(51) - f(49)}{51-49} = f'(c)$$ where$$f'(c) = \frac{1}{2 \sqrt{c}} < \frac{1}{2 \sqrt{49}} = \frac 1{14}$$ and$$f'(c) = \frac{1}{2 \sqrt{c}} > \frac{1}{2 \sqrt{64}} = \frac 1{16}.$$Thus$$\frac 1{16} < \frac{\sqrt{51} - 7}{2} < \frac 1{14}.$$
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