Final answer:
To find the number of relative extrema for f(x) with f'(x) = sin(x² +1) on the interval 2 < x < 4, we look for sign changes in the derivative that imply extrema. The sine function's behavior indicates potential for extrema when x² +1 is a multiple of π, but precise computation or graphing is needed to confirm the exact number of relative extrema in this interval.
Step-by-step explanation:
To determine how many relative extrema the function f with derivative f ' (x) = sin (x² +1) has on the interval 2 < x < 4, we will investigate where the derivative changes sign, as this indicates the potential for a relative maximum or minimum.
First, since the sine function oscillates between -1 and 1, f ' (x) will change sign whenever x² + 1 equates to an angle where the sine function is zero; these are the multiples of π (pi). So we are looking for values of x for which x² +1 is a multiple of π.
Since the derivative does not have a simple algebraic expression for its zeros, we would typically need to use a graphing calculator or a numerical method to determine precisely where the derivative changes sign. However, since the interval is from 2 to 4, we can reason about the rough location of these sign changes. The sine function has zeroes at multiples of π, but none between x = √(π - 1) and x = √(2π - 1) are present in our interval of interest. Hence, within the interval 2 < x < 4, it's possible that there are no sign changes implying no relative extrema.
However, without the actual computation of the roots, we cannot definitively answer how many relative extrema exist. The investigation would require numerical methods or graphing technology to find the exact number of times the function sin (x² +1) crosses the x-axis within the given interval to determine the number of relative extrema.