The correct answer is: C. only a maxima at x = 1
The function $f(x) = 2x – x^2 + 3$ is a parabola. The parabola opens upwards, so it has a maximum point. The maximum point occurs when $x = 1$.
To find the maximum point, we can take the derivative of $f(x)$ and set it equal to zero. The derivative of $f(x)$ is $f'(x) = 2 – 2x$. Setting $f'(x) = 0$, we get $2 – 2x = 0$. Solving for $x$, we get $x = 1$.
To verify that $x = 1$ is a maximum point, we can evaluate $f'(x)$ at $x = 1$. We get $f'(1) = 2 – 2(1) = 0$. Since $f'(1) = 0$, we know that $x = 1$ is a critical point. The derivative of a function is always zero at a critical point. If the derivative is positive on one side of the critical point and negative on the other side, then the critical point is a maximum point. In this case, the derivative of $f(x)$ is positive to the left of $x = 1$ and negative to the right of $x = 1$. Therefore, $x = 1$ is a maximum point.
The other options are incorrect because they state that the function has a minimum point at $x = 1$. However, the function does not have a minimum point at $x = 1$. The function has a maximum point at $x = 1$.