Final answer:
For the function f(x) = x² + 1, the derivative f'(x) = 2x is zero at x = 0, and the second derivative f''(x) = 2 is positive, indicating a relative minimum at (0,1).
Step-by-step explanation:
To find the relative extrema of a function, we often take the derivative of the function and find where it equals zero. This indicates potential points of relative maxima or minima. If the second derivative at that point is positive, we have a relative minimum; if it is negative, we have a relative maximum. The function f(x) = x² + 1 - 6 seems to be a misprint, but assuming the correct form is f(x) = x² + 1, we can find the first derivative as f'(x) = 2x. Setting this equal to zero, we find x = 0. To ascertain the nature of this extrema, we would take the second derivative, which is f''(x) = 2, a positive number. Hence, f(x) has a relative minimum at (0,1). No option given matches this result, indicating either the function provided is incorrect or the options are incorrect.