1. We are given the derivative of a function: $f'(x) = x^2 - 3 - 3 \sin(2x^2 - x)$ and the domain $x \in [0,4]$. 2. The problem is to understand or analyze this derivative function over the interval $[0,4]$. 3. The derivative $f'(x)$ tells us the rate of change of the original function $f(x)$ at any point $x$. 4. The function $f'(x)$ is composed of a polynomial part $x^2 - 3$ and a trigonometric part $-3 \sin(2x^2 - x)$. 5. To analyze $f'(x)$, we can look for critical points where $f'(x) = 0$, which correspond to potential maxima, minima, or inflection points of $f(x)$. 6. Setting $f'(x) = 0$ gives: $$x^2 - 3 - 3 \sin(2x^2 - x) = 0$$ 7. This is a transcendental equation and generally requires numerical methods to solve for $x$ in $[0,4]$. 8. We can also analyze the behavior of $f'(x)$ by evaluating it at key points: - At $x=0$: $f'(0) = 0^2 - 3 - 3 \sin(0) = -3$ - At $x=4$: $f'(4) = 16 - 3 - 3 \sin(2\cdot16 - 4) = 13 - 3 \sin(28)$ 9. Since $\sin$ oscillates between $-1$ and $1$, the term $-3 \sin(2x^2 - x)$ oscillates between $-3$ and $3$, affecting the shape of $f'(x)$. 10. To fully understand $f(x)$, one would integrate $f'(x)$ over $[0,4]$ or analyze $f'(x)$ graphically. Final answer: The derivative function is $f'(x) = x^2 - 3 - 3 \sin(2x^2 - x)$ for $x \in [0,4]$, which combines polynomial growth and oscillations due to the sine term.