1. **Problem Statement:** Find the maximum, minimum, relative maximum, and relative minimum values of the function $$f(x) = \sin(x^2)$$ over the interval $$(0, 4\pi)$$. 2. **Formula and Rules:** To find extrema, we first find critical points by solving $$f'(x) = 0$$ where $$f'(x)$$ is the derivative of $$f(x)$$. 3. **Derivative Calculation:** Using the chain rule, $$f'(x) = \cos(x^2) \cdot 2x = 2x \cos(x^2)$$. 4. **Find Critical Points:** Set $$f'(x) = 0$$: $$2x \cos(x^2) = 0$$ This implies either $$x = 0$$ or $$\cos(x^2) = 0$$. Since the interval is $$(0, 4\pi)$$, exclude $$x=0$$. 5. **Solve $$\cos(x^2) = 0$$:** $$\cos(\theta) = 0$$ at $$\theta = \frac{\pi}{2} + k\pi$$ for integers $$k$$. Set $$x^2 = \frac{\pi}{2} + k\pi$$. 6. **Find $$x$$ values:** $$x = \sqrt{\frac{\pi}{2} + k\pi}$$ for integers $$k$$ such that $$0 < x < 4\pi$$. 7. **Determine $$k$$ range:** Since $$x < 4\pi \approx 12.566$$, $$x^2 < (4\pi)^2 = 16\pi^2 \approx 157.91$$. So, $$\frac{\pi}{2} + k\pi < 157.91$$ $$k < \frac{157.91 - \frac{\pi}{2}}{\pi} \approx \frac{157.91 - 1.57}{3.14} \approx 49.5$$. So $$k = 0,1,2,...,49$$. 8. **Evaluate $$f(x)$$ at critical points:** At each $$x_k = \sqrt{\frac{\pi}{2} + k\pi}$$, $$f(x_k) = \sin(x_k^2) = \sin\left(\frac{\pi}{2} + k\pi\right)$$. Since $$\sin\left(\frac{\pi}{2} + k\pi\right) = (-1)^k$$, - For even $$k$$, $$f(x_k) = 1$$ (relative maxima). - For odd $$k$$, $$f(x_k) = -1$$ (relative minima). 9. **Check endpoints:** At $$x=0$$ (excluded), at $$x=4\pi$$, $$f(4\pi) = \sin((4\pi)^2) = \sin(16\pi^2)$$. Since $$16\pi^2$$ is not a multiple of $$\pi$$, approximate: $$16\pi^2 \approx 157.91$$, $$157.91 \mod 2\pi \approx 157.91 - 25 \times 2\pi = 157.91 - 157.08 = 0.83$$, so $$f(4\pi) \approx \sin(0.83) \approx 0.74$$. 10. **Summary:** - Relative maxima at $$x = \sqrt{\frac{\pi}{2} + 2m\pi}$$ with value 1. - Relative minima at $$x = \sqrt{\frac{\pi}{2} + (2m+1)\pi}$$ with value -1. - Absolute maximum value is 1. - Absolute minimum value is -1. **Final answer:** - Maximum value: $$1$$ - Minimum value: $$-1$$ - Relative maxima at $$x = \sqrt{\frac{\pi}{2} + 2m\pi}$$ for integers $$m$$ with $$0 \leq m \leq 24$$. - Relative minima at $$x = \sqrt{\frac{\pi}{2} + (2m+1)\pi}$$ for integers $$m$$ with $$0 \leq m \leq 24$$.