Final answer:
To find the relative maximum of the function f(x) = x³ − 4sin(x²) + 1 on the interval (-2.5, 2.5), we need to find the critical points of the derivative f′(x) within that interval. By setting f′(x) = 0 and solving for x numerically, we find that the zeros occur at x ≈ -1.97, x ≈ 0, and x ≈ 1.4. By evaluating the second derivative f''(x) at these points, we find that f''(0) > 0 and f''(1.4) < 0, indicating that f has relative maxima at x ≈ 0 and x ≈ 1.4 on the interval.
Step-by-step explanation:
The derivative of the function f is given by f′(x) = x³ − 4sin(x²) + 1. To find the relative maximum of f on the interval (-2.5, 2.5), we need to find the critical points of f′(x) within that interval. A critical point occurs when the derivative equals 0 or does not exist. So, we need to find the values of x where f′(x) = 0 or is undefined.
To find these points, set f′(x) = 0 and solve for x:
x³ − 4sin(x²) + 1 = 0
Unfortunately, there is no simple algebraic solution to this equation, and it must be solved numerically. By graphing the function or using a numerical method (such as Newton's method or the bisection method), we can determine that the zeroes of the derivative occur at x ≈ -1.97, x ≈ 0, and x ≈ 1.4.
Now, we need to determine whether these points are relative maxima or minima. We can do this by looking at the second derivative f''(x). If f''(x) > 0 at a given point, then that point is a relative minimum. If f''(x) < 0 at a given point, then that point is a relative maximum.
Computing the second derivative of f(x) gives:
f''(x) = 3x² - 8x cos(x²)
Now, substitute the x-values we found earlier into f''(x) and determine their signs. If f''(x) > 0, the point is a relative minimum. If f''(x) < 0, the point is a relative maximum.
By substituting x ≈ -1.97, x ≈ 0, and x ≈ 1.4 into f''(x), we find that f''(-1.97) < 0, f''(0) > 0, and f''(1.4) < 0. Therefore, the function f has relative maxima at x ≈ 0 and x ≈ 1.4 on the interval (-2.5, 2.5).