Given the function f, of, x, equals, integral, from, 1, to, x, of, left bracket, t, to the power 4 , minus, 30, t, squared, right bracket, d, t, commaf(x)=∫ 1x (t 4 −30t 2 )dt, determine all xx-values, if any, for which ff has a relative maximum.
Question
Given the function f, of, x, equals, integral, from, 1, to, x, of, left bracket, t, to the power 4 , minus, 30, t, squared, right bracket, d, t, commaf(x)=∫ 1x (t 4 −30t 2 )dt, determine all xx-values, if any, for which ff has a relative maximum.
Solution 1
To find the x-values for which f has a relative maximum, we first need to find the derivative of f. The derivative of an integral from a constant to x of a function is just the function evaluated at x. So, the derivative f'(x) of f(x) is x^4 - 30x^2.
Next, we set the derivative equal to zero and solve for x to find critical points:
x^4 - 30x^2 = 0 x^2(x^2 - 30) = 0 x^2 = 0 or x^2 = 30 x = 0 or x = ±sqrt(30)
So, the critical points are x = 0, x = sqrt(30), and x = -sqrt(30).
To determine whether these are relative maxima, we need to check the second derivative at these points. The second derivative f''(x) is 4x^3 - 60x.
f''(0) = 0 f''(sqrt(30)) = 4(sqrt(30))^3 - 60sqrt(30) = 120sqrt(30) - 60sqrt(30) = 60sqrt(30) > 0 f''(-sqrt(30)) = 4(-sqrt(30))^3 - 60(-sqrt(30)) = -120sqrt(30) + 60sqrt(30) = -60sqrt(30) < 0
So, by the second derivative test, f has a relative maximum at x = -sqrt(30) and a relative minimum at x = sqrt(30). The second derivative at x = 0 does not give us any information, so we cannot conclude whether f has a relative maximum, minimum, or neither at x = 0.
Solution 2
To find the x-values for which f has a relative maximum, we first need to find the derivative of f. The derivative of an integral from a constant to x of a function is just the function evaluated at x. So, the derivative f'(x) of f(x) is x^4 - 30x^2.
Next, we set the derivative equal to zero and solve for x to find critical points:
x^4 - 30x^2 = 0 x^2(x^2 - 30) = 0 x^2 = 0 or x^2 = 30 x = 0 or x = ±sqrt(30)
So, the critical points are x = 0, x = sqrt(30), and x = -sqrt(30).
To determine whether these are relative maxima, we need to check the second derivative at these points. The second derivative f''(x) is 4x^3 - 60x.
f''(0) = 0 f''(sqrt(30)) = 4(sqrt(30))^3 - 60sqrt(30) = 120sqrt(30) - 60sqrt(30) = 60sqrt(30) > 0 f''(-sqrt(30)) = 4(-sqrt(30))^3 - 60(-sqrt(30)) = -120sqrt(30) + 60sqrt(30) = -60sqrt(30) < 0
So, by the second derivative test, f has a relative maximum at x = -sqrt(30) and a relative minimum at x = sqrt(30). The second derivative at x = 0 does not give us any information, so we cannot conclude whether f has a relative maximum, minimum, or neither at x = 0.
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