1. **Problem:** Given the function $f(x) = \sin(2x + 1)$ for $0 \leq x \leq \pi$, find the $x$-coordinates of the maximum and minimum points of $f(x)$, correct to one decimal place. 2. **Formula and rules:** The maximum and minimum points of a sine function occur where its derivative is zero and changes sign. The derivative of $f(x)$ is: $$f'(x) = 2 \cos(2x + 1)$$ Set $f'(x) = 0$ to find critical points: $$2 \cos(2x + 1) = 0 \implies \cos(2x + 1) = 0$$ 3. **Solve for critical points:** $$\cos(\theta) = 0 \implies \theta = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ Let $\theta = 2x + 1$, so: $$2x + 1 = \frac{\pi}{2} + k\pi$$ Solve for $x$: $$x = \frac{\frac{\pi}{2} + k\pi - 1}{2}$$ 4. **Find values of $x$ in $[0, \pi]$:** For $k=0$: $$x = \frac{\frac{\pi}{2} - 1}{2} = \frac{1.5708 - 1}{2} = \frac{0.5708}{2} = 0.2854$$ For $k=1$: $$x = \frac{\frac{\pi}{2} + \pi - 1}{2} = \frac{1.5708 + 3.1416 - 1}{2} = \frac{3.7124}{2} = 1.8562$$ For $k=2$: $$x = \frac{\frac{\pi}{2} + 2\pi - 1}{2} = \frac{1.5708 + 6.2832 - 1}{2} = \frac{6.854}{2} = 3.427$$ Since $\pi \approx 3.1416$, $x=3.427$ is outside the interval $[0, \pi]$, so discard. 5. **Determine maxima and minima:** Evaluate second derivative: $$f''(x) = -4 \sin(2x + 1)$$ At $x=0.2854$: $$f''(0.2854) = -4 \sin(2(0.2854) + 1) = -4 \sin(1.5708) = -4 \times 1 = -4 < 0$$ So $x=0.3$ (to 1 decimal place) is a maximum. At $x=1.8562$: $$f''(1.8562) = -4 \sin(2(1.8562) + 1) = -4 \sin(4.7124) = -4 \times (-1) = 4 > 0$$ So $x=1.9$ is a minimum. **Final answer:** - Maximum at $x = 0.3$ - Minimum at $x = 1.9$