Final answer:
The value of the derivative at the extremum for the function f(x) = 2x² + 3 is zero since the first derivative, f'(x) = 4x, equals zero at x = 0.
Step-by-step explanation:
To find the value of the derivative at the indicated extremum for the function f(x) = x² + x² + 3, you first simplify the function as f(x) = 2x² + 3. Then, take the first derivative of f(x) with respect to x, which is f'(x) = 4x. At an extremum, the derivative is zero, so set f'(x) = 0 and solve for x. In this case, you get x = 0. Since the function is a parabola opening upwards (as indicated by the positive coefficient of x²), x = 0 corresponds to the vertex and is the extremum, which is a minimum in this case. The value of the derivative at x = 0 is f'(0) = 4(0) = 0.