1. **State the problem:** We need to verify if the function $f(x) = 1 - \cos\left(x^{\frac{1}{2}}\right)$ satisfies the conditions of Rolle's theorem on the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. 2. **Recall Rolle's theorem:** If a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$, then there exists at least one $c \in (a,b)$ such that $f'(c) = 0$. 3. **Check continuity and differentiability:** - The function involves $\cos\left(x^{\frac{1}{2}}\right)$. The square root $x^{\frac{1}{2}}$ is only defined for $x \geq 0$ in the real numbers, but the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ includes negative values. - Therefore, $f(x)$ is not defined for negative $x$ in the interval, so it is not continuous on the entire interval. 4. **Conclusion:** Since $f$ is not continuous on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, Rolle's theorem does not apply. **If the problem intended $f(x) = 1 - \cos\left(\sqrt{x}\right)$ on $[0, \frac{\pi}{2}]$, then:** 5. **Check $f(a)$ and $f(b)$:** - $f(0) = 1 - \cos(0) = 1 - 1 = 0$ - $f\left(\frac{\pi}{2}\right) = 1 - \cos\left(\sqrt{\frac{\pi}{2}}\right)$ (not necessarily zero) - Since $f(0) \neq f\left(\frac{\pi}{2}\right)$, Rolle's theorem does not apply. 6. **Derivative:** $$f'(x) = -\left(-\sin\left(x^{\frac{1}{2}}\right)\right) \cdot \frac{1}{2} x^{-\frac{1}{2}} = \frac{\sin\left(\sqrt{x}\right)}{2 \sqrt{x}}$$ 7. **Find $c$ such that $f'(c) = 0$:** - $\sin\left(\sqrt{c}\right) = 0$ - $\sqrt{c} = k\pi$, $k \in \mathbb{Z}$ - Since $c \in (0, \frac{\pi}{2})$, the only solution is $c=0$, which is not in the open interval. **Final answer:** Rolle's theorem does not apply to the given function on the specified interval because the function is not continuous on the interval and $f(a) \neq f(b)$.