Question: Apply the Mean Value Theorem to $f(x) = 2^x = e^{x\ln 2}$ , show that $\frac{1}{2}\ln 2 < \sqrt{2} - 1$.
My Attempt:
According to the Mean Value Theorem:
$$f' (c) = \frac{f(b) - f(a)}{b - a}$$ Where $a < b < c$
To apply the Mean Value Theorem, we first note that $f(x)$ is continuous on the interval $[0,\sqrt{2}]$ and differentiable on $(0,\sqrt{2})$. Thus, there exists a number $c\in (0,\sqrt{2})$ such that
$$f' (c) = \frac{f(\sqrt{2}) - f(0)}{\sqrt{2} - 0}$$
Taking derivatives, we have
$$f'(x) = (\ln 2)2^x = (\ln 2)e^{x\ln 2}$$
Substituting these values into the above equation, we get
$$(\ln 2)2^c = \frac{2^{\sqrt{2}} - 1}{\sqrt{2}}$$
Solving for $c$, we obtain
$$c = \frac{\ln\left(\frac{2^{\sqrt{2}} - 1}{\sqrt{2}\ln 2}\right)}{\ln 2}$$
Now, we want to show that $\frac{1}{2}\ln 2 < \sqrt{2} - 1$. We start by simplifying the right-hand side:
\begin{align*} \sqrt{2} - 1 &= \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} + 1}{\sqrt{2} + 1} \ &= \frac{(\sqrt{2} - 1)(\sqrt{2} + 1)}{2} \ &= \frac{1}{2}(2 - 1) \ &= \frac{1}{2}. \end{align*}
Thus, we need to show that $\frac{1}{2}\ln 2 < \frac{1}{2}$. This is equivalent to showing that $\ln 2 < 1$, which is true since $e^1 = e > 2$.
Therefore, we have shown that $\frac{1}{2}\ln 2 < \sqrt{2} - 1$ using the Mean Value Theorem.
Updated Attempt:
Using the Mean Value Theorem on $f(x)=2^x$ on the interval $[0, \frac{1}{2}]$, we have:
$$f' (c) = \frac{f(\frac{1}{2}) - f(0)}{\frac{1}{2} - 0} = \frac{\sqrt{2} - 1}{\frac{1}{2}} = 2\sqrt{2} - 2$$
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:
$$\frac{16}{7} > (\ln 2)^{-2}$$
Am I right ?
Hint: Apply the MVT to the function $f(x) = 2^x$ on the interval $\left[0,\dfrac{1}{2}\right]$.