Use the Mean Value Theorem to prove an Inequality.

Question

Use the Mean Value Theorem to show that $$\frac{1}{9}<\sqrt{66}-8<\frac{1}{8}$$

I really don't know how to use the mean value theorem for this as $$\frac{d}{dx}\sqrt{x}\frac{1}{2\sqrt{x}}$$ I would give more but I'm really stuck. Any tips would be greatly appreciated

Answer 1

$\sqrt{x}$ satisfies all the conditions of MVT on [64, 66]. So there exists some c ∈ (64,66) such that $\frac{\sqrt{66} - \sqrt{64}}{66 - 64}$ = $\frac{1}{2\sqrt{c}}$. So $\sqrt{66} - 8 = \frac{1}{\sqrt{c}}$. We know $\frac{1}{\sqrt{x}}$ is a strictly decreasing function on (64, 66), thus for any $n ∈ (64, 66)$, $\frac{1}{\sqrt{66}}$ < $\frac{1}{\sqrt{n}}$ < $\frac{1}{\sqrt{64}} = \frac{1}{8}.$ You can combine these, and that $\frac{1}{\sqrt{66}}$ > $\frac{1}{\sqrt{81}}$ = $\frac{1}{9}$, to finish the proof - can you see how?

Answer 2

Not using MVT, you have between 64 and 81 the derivative is monotonically decreasing from $\frac 1 {16}$ to $\frac 1 {18}$.

This means taking a linear approximation at $x=64$, for values in $(64,81)$ $$\frac 1 {18}(x-64)+8< \sqrt {x}<\frac 1 {16}(x-64)+8$$ plugging in $x=66$ gets you there. Not sure where MVT is supposed to come into this