Let g be the function defined by g(x)=∫x−1(−12+cos(t3+2t))ⅆt for 0
0.471
0.471
To find the relative maximum of g(x), we need to find the critical points of g(x) and then check which one of these points corresponds to the relative maximum.
Using the Fundamental Theorem of Calculus, we can differentiate g(x) to find its derivative:
g'(x) = (-1/2 + cos(x^3 + 2x))/x
To find the critical points, we need to set g'(x) equal to zero and solve for x:
(-1/2 + cos(x^3 + 2x))/x = 0
cos(x^3 + 2x) = -1/2
x^3 + 2x = 2nπ ± 2π/3, where n is an integer.
Since the given interval is 0 < x < π/2, we only need to consider the positive root of x^3 + 2x = 2nπ + 2π/3. After plotting the graph of y = x^3 + 2x - 2π/3, we can see that it intersects the x-axis only once. We can use numerical methods like Newton's method or bisection method to find the root, which is approximately 0.6283. Now, we need to check whether this critical point corresponds to a relative maximum or not. We can do this by examining the sign of g''(x) at x = 0.6283. Differentiating g'(x), we get: g''(x) = [x(cos(x^3+2x))+3sin(x^3+2x)+1/2]/x^2 Evaluating g''(0.6283), we get: g''(0.6283) ≈ 0.6243 Since g''(0.6283) is positive, we can conclude that x = 0.6283 is a point of relative minimum. Therefore, g(x) attains a relative maximum at x = 0 (since g(x) is continuous and increasing on the interval [0, 0.6283] and then decreasing on the interval [0.6283, π/2]).
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