Locate and classify the critical point using the Second Derivative Test

First, differentiate: $f'(x) = \frac{1}{2\sqrt{x}} - 1$. Setting $f'(x) = 0$ gives $\frac{1}{2\sqrt{x}} = 1$, so $\sqrt{x} = \frac{1}{2}$, and thus $x = \frac{1}{4}$. This value lies in $(0,4)$, so it is the only critical number to consider (note $f'(x)$ is undefined at $x=0$, but that's an endpoint, not an interior critical point for this test).

Next, compute the second derivative: $f''(x) = -\frac{1}{4}x^{-3/2}$. Since $x = \frac{1}{4} > 0$, we get $f''(1/4) = -\frac{1}{4}(1/4)^{-3/2} = -\frac{1}{4}(8) = -2 < 0$. Because $f''(1/4) < 0$, the Second Derivative Test confirms $x = \frac{1}{4}$ is a local maximum. Evaluating $f(1/4) = \sqrt{1/4} - 1/4 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.