1. We are given the derivative of a function: $f'(x)=\cos \left( \frac{x}{2}\right) -3\sin (x^2)$. We need to find the value of $x$ in the interval $(0,2)$ where the graph of $f$ has a point of inflection. 2. A point of inflection occurs where the second derivative $f''(x)$ changes sign, which means we need to find where $f''(x)=0$. 3. First, find $f''(x)$ by differentiating $f'(x)$: $$f''(x) = \frac{d}{dx} \left( \cos \left( \frac{x}{2} \right) - 3 \sin (x^2) \right)$$ 4. Differentiate each term: - For $\cos \left( \frac{x}{2} \right)$, use the chain rule: $$\frac{d}{dx} \cos \left( \frac{x}{2} \right) = -\sin \left( \frac{x}{2} \right) \cdot \frac{1}{2} = -\frac{1}{2} \sin \left( \frac{x}{2} \right)$$ - For $-3 \sin (x^2)$, use the chain rule: $$\frac{d}{dx} (-3 \sin (x^2)) = -3 \cos (x^2) \cdot 2x = -6x \cos (x^2)$$ 5. So, $$f''(x) = -\frac{1}{2} \sin \left( \frac{x}{2} \right) - 6x \cos (x^2)$$ 6. Set $f''(x) = 0$ to find possible inflection points: $$-\frac{1}{2} \sin \left( \frac{x}{2} \right) - 6x \cos (x^2) = 0$$ 7. Rearranged: $$-\frac{1}{2} \sin \left( \frac{x}{2} \right) = 6x \cos (x^2)$$ Multiply both sides by $-2$: $$\cancel{-2} \cdot \left(-\frac{1}{2} \sin \left( \frac{x}{2} \right) \right) = \cancel{-2} \cdot 6x \cos (x^2)$$ $$\sin \left( \frac{x}{2} \right) = -12x \cos (x^2)$$ 8. We need to find $x$ in $(0,2)$ satisfying: $$\sin \left( \frac{x}{2} \right) + 12x \cos (x^2) = 0$$ 9. This transcendental equation cannot be solved exactly by algebraic methods, so numerical methods (like Newton-Raphson) or graphing are used. 10. By testing values or graphing, the solution in $(0,2)$ is approximately $x \approx 0.1$ (more precise numerical methods can refine this). Hence, the point of inflection occurs near $x \approx 0.1$ in the interval $(0,2)$.