Question:
Apply the Mean Value Theorem to $f(x) = 2^x = e^{x\ln 2}$ , show that $\frac{1}{2}\ln 2 < \sqrt{2} - 1$.
Attempt:
Using the Mean Value Theorem on $f(x)=2^x$ on the interval $[0, \frac{1}{2}]$, we have:
$$f' (c) = \frac{f(\frac{1}{2}) - f(0)}{\frac{1}{2} - 0} = \frac{\sqrt{2} - 1}{\frac{1}{2}} = 2\sqrt{2} - 2$$
Therefore, we have:
$$2\sqrt{2} - 2 = f' (c) = (\ln 2)2^c$$
Solving for $c$, we get:
$$c = \frac{1}{2}\ln(2\sqrt{2}-2) < \frac{3}{4}\ln 2$$
Now, we just need to show that $\frac{3}{4}\ln 2 < \sqrt{2} - 1$. Squaring both sides, we get:
$$\frac{9}{16}\ln^2 2 < 3 - 2\sqrt{2}$$
Subtracting $3$ from both sides, we have:
$$-\frac{9}{16}\ln^2 - 1 < -2\sqrt{2}$$
Dividing both sides by $-2\sqrt{2}$, we get:
$$\frac{7}{8\sqrt{2}}\ln^2 2 > 1$$
Finally, we can simplify the left-hand side to get:
$$\frac{7}{16}(\ln 2)^2 > 1$$
Multiplying both sides by $\frac{16}{7}(\frac{1}{\ln 2})^2$, we get:
$$\frac{16}{7} > (\ln 2)^{-2}$$
Updated Attempt:
Using the Mean Value Theorem on $f(x)=2^x$ on the interval $[0, \frac{1}{2}]$, we have:
$$f' (c) = \frac{f(\frac{1}{2}) - f(0)}{\frac{1}{2} - 0} = \frac{\sqrt{2} - 1}{\frac{1}{2}}$$
$f$ is an increasing function since $f' > 0$. Clearly $f'$ is also increasing. Therefore, for $c \in (0, \frac{1}{2})$, we have:
$$f'(c) < f'(\frac{1}{2}) = \ln 2$$
So, we have:
$$\frac{\sqrt{2} - 1}{\frac{1}{2}} < \ln 2$$
Simplifying the left-hand side gives:
$$\frac{\sqrt{2} - 1}{\frac{1}{2}} = 2\sqrt{2} - 2$$
Substituting back in the inequality gives:
$$2\sqrt{2} - 2 < \ln 2$$
Adding 2 to both sides yields:
$$2\sqrt{2} < \ln 2 + 2$$
Dividing both sides by 2 gives:
$$\sqrt{2} < \frac{1}{2}\ln 2 + 1$$
Finally, we get:
$$\frac{1}{2}\ln 2 < \sqrt{2} - 1$$
So, we have shown that $\frac{1}{2}\ln 2 < \sqrt{2} - 1$.
Am I right ?
Using the Mean Value Theorem on $f(x)=2^x$ on the interval $[0, \frac{1}{2}]$, we have:
$$f' (c) = \frac{f(\frac{1}{2}) - f(0)}{\frac{1}{2} - 0} = \frac{\sqrt{2} - 1}{\frac{1}{2}}$$
$f$ is an increasing function since $f' > 0$. Clearly $f'$ is also increasing. Therefore $f'(c) > f'(0) = \ln 2$, which completes the proof.