1. The problem is to analyze the function $f(x) = 1 - \cos\left(x^{\frac{1}{2}}\right)$ on the interval $(a,b)$.\n\n2. The function involves the cosine of the square root of $x$. We need to understand its behavior, domain, and range.\n\n3. The domain of $f(x)$ is $x \geq 0$ because $x^{\frac{1}{2}} = \sqrt{x}$ is defined for non-negative $x$.\n\n4. The cosine function $\cos(\theta)$ has values between $-1$ and $1$. Therefore, $1 - \cos(\sqrt{x})$ ranges between $1 - 1 = 0$ and $1 - (-1) = 2$.\n\n5. To find critical points, we differentiate $f(x)$:\n$$f'(x) = -\frac{d}{dx} \cos(\sqrt{x}) = -(-\sin(\sqrt{x})) \cdot \frac{1}{2\sqrt{x}} = \frac{\sin(\sqrt{x})}{2\sqrt{x}}$$\nfor $x > 0$.\n\n6. The derivative $f'(x) = \frac{\sin(\sqrt{x})}{2\sqrt{x}}$ is zero when $\sin(\sqrt{x}) = 0$, i.e., when $\sqrt{x} = n\pi$ for $n=0,1,2,\ldots$. Thus, critical points occur at $x = n^2 \pi^2$.\n\n7. At these points, $f(x) = 1 - \cos(n\pi) = 1 - (-1)^n$. For even $n$, $f(x) = 0$ (minimum), and for odd $n$, $f(x) = 2$ (maximum).\n\n8. The function oscillates between 0 and 2 as $x$ increases, with minima at $x = (2k)^2 \pi^2$ and maxima at $x = (2k+1)^2 \pi^2$ for integers $k \geq 0$.\n\nFinal answer: The function $f(x) = 1 - \cos(\sqrt{x})$ is defined for $x \geq 0$, oscillates between 0 and 2, with critical points at $x = n^2 \pi^2$ where minima are 0 for even $n$ and maxima are 2 for odd $n$.