1. The problem states that the derivative of a function $f$ is given by $f'(x) = x^4 + 3x^2 - 2$ and that $f(0) = 4$. We need to find the value of $f(1)$. 2. To find $f(1)$, we first need to find the original function $f(x)$ by integrating $f'(x)$. The formula for the antiderivative is: $$f(x) = \int f'(x) \, dx + C$$ where $C$ is the constant of integration. 3. Integrate each term of $f'(x)$: $$\int x^4 \, dx = \frac{x^5}{5}$$ $$\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3$$ $$\int -2 \, dx = -2x$$ So, $$f(x) = \frac{x^5}{5} + x^3 - 2x + C$$ 4. Use the initial condition $f(0) = 4$ to find $C$: $$f(0) = \frac{0^5}{5} + 0^3 - 2 \cdot 0 + C = C = 4$$ So, $C = 4$. 5. The function is: $$f(x) = \frac{x^5}{5} + x^3 - 2x + 4$$ 6. Now find $f(1)$: $$f(1) = \frac{1^5}{5} + 1^3 - 2 \cdot 1 + 4 = \frac{1}{5} + 1 - 2 + 4 = \frac{1}{5} + 3 = \frac{1}{5} + \frac{15}{5} = \frac{16}{5}$$ 7. Therefore, $f(1) = \frac{16}{5} = 3.2$.