Question

Let \(0

Short answer

When applying the Mean Value Theorem on the continuous and differentiable function \(f(x) = \sqrt{x}\) on the interval [a,b], it can be proven that \(\sqrt{b}-\sqrt{a}<\frac{b-a}{2 \sqrt{a}}\). The key was to observe that \(c>a\) and hence, \(\sqrt{c}>\sqrt{a}\). Therefore, \(\frac{1}{2\sqrt{c}}<\frac{1}{2\sqrt{a}}\) which in turn led to the required result.

Step-by-step solution

Define the function

Choose the function \(f(x) = \sqrt{x}\), because we have two square roots in the inequality. This function is continuous on the interval [a, b] and differentiable on the open interval (a, b), therefore the conditions of the Mean Value Theorem apply.

Apply the Mean Value Theorem

The Mean Value Theorem states that there exists some \(c \in (a, b)\) such that \(f'(c)= \frac{f(b)-f(a)}{b-a}\). If you compute this, you find \(f'(c) = \frac{\sqrt{b} - \sqrt{a}}{b-a}\). Therefore, there must be some \(c \in (a, b)\) such that this is true.

Find \(f'(c)\)

We find \(f'(x) = \frac{1}{2\sqrt{x}}\) by the derivative rule. Substitute \(x = c\) into this equation, we get \(f'(c) = \frac{1}{2\sqrt{c}}\).

Relate \(f'(c)\) to \(\frac{\sqrt{b}-\sqrt{a}}{b - a}\)

Therefore, \(\frac{\sqrt{b}-\sqrt{a}}{b - a} = \frac{1}{2\sqrt{c}}\). As \(c \in (a, b)\), we have \(c > a\), that simply means \(\sqrt{c} > \sqrt{a}\) and hence, \(\frac{1}{2\sqrt{c}} < \frac{1}{2\sqrt{a}}\). This means \(\frac{\sqrt{b}-\sqrt{a}}{b - a} < \frac{1}{2\sqrt{a}}\) which rearranges to \(\sqrt{b}-\sqrt{a} < \frac{b - a}{2\sqrt{a}}\)