Let g be the function defined by g(x)=∫x−1(−12+cos(t3+2t))ⅆt for 0
0.471
To find the value of x at which g attains a relative maximum, we need to use the First Derivative Test.
1. Calculate g'(x):
Using the Fundamental Theorem of Calculus, we can find the derivative of g(x) as follows:
g'(x) = [-1/2 + cos(x^3 + 2x)] / (x^(-1))
Multiplying by x^2 on both numerator and denominator, we get:
g'(x) = [-x^2/2 + x^2cos(x^3+2x)] / 1
Simplifying further:
g'(x) = x^2[cos(x^3+2x) – 1/2]
2. Find critical points:
To find critical points, we need to solve the equation g'(x) = 0:
x^2[cos(x^3+2x) – 1/2] = 0
Therefore, either x = 0 or cos(x^3+2x) = 1/2.
For x = 0, g is not defined.
For cos(x^3+2x) = 1/2, we can use the fact that cos(π/3) = 1/2, to get:
x^3 + 2x = π/3
This equation cannot be solved analytically, but we can use numerical methods or a graphing calculator to find the approximate solution: x ≈ 0.715.
Thus, the critical point of g is x = 0.715.
3. Use the First Derivative Test:
To determine if x = 0.715 is a relative maximum, we check the sign of g'(x) on either side of the critical point:
For x < 0.715, x^2 is positive, and cos(x^3+2x) is negative. Therefore, g'(x) is negative.
For x > 0.715, x^2 is positive, and cos(x^3+2x) is positive. Therefore, g'(x) is positive.
Thus, by the First Derivative Test, x = 0.715 is a relative maximum of g.
Therefore, the value of x at which g attains a relative maximum is x = 0.715.
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0.471
To find the value of x at which g attains a relative maximum, we need to use the First Derivative Test.
1. Calculate g'(x):
Using the Fundamental Theorem of Calculus, we can find the derivative of g(x) as follows:
g'(x) = [-1/2 + cos(x^3 + 2x)] / (x^(-1))
Multiplying by x^2 on both numerator and denominator, we get:
g'(x) = [-x^2/2 + x^2cos(x^3+2x)] / 1
Simplifying further:
g'(x) = x^2[cos(x^3+2x) – 1/2]
2. Find critical points:
To find critical points, we need to solve the equation g'(x) = 0:
x^2[cos(x^3+2x) – 1/2] = 0
Therefore, either x = 0 or cos(x^3+2x) = 1/2.
For x = 0, g is not defined.
For cos(x^3+2x) = 1/2, we can use the fact that cos(π/3) = 1/2, to get:
x^3 + 2x = π/3
This equation cannot be solved analytically, but we can use numerical methods or a graphing calculator to find the approximate solution: x ≈ 0.715.
Thus, the critical point of g is x = 0.715.
3. Use the First Derivative Test:
To determine if x = 0.715 is a relative maximum, we check the sign of g'(x) on either side of the critical point:
For x < 0.715, x^2 is positive, and cos(x^3+2x) is negative. Therefore, g'(x) is negative. For x > 0.715, x^2 is positive, and cos(x^3+2x) is positive. Therefore, g'(x) is positive.
Thus, by the First Derivative Test, x = 0.715 is a relative maximum of g.
Therefore, the value of x at which g attains a relative maximum is x = 0.715.
More Answers:
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