To find the critical points of the function f(x) = sin^3(x)cos(x), we need to find the values of x where the derivative of f(x) is equal to zero or does not exist.
First, we find the derivative of f(x) using the chain rule:
f'(x) = 3sin^2(x)cos^2(x) - sin^4(x)
To find the critical points, we set the derivative equal to zero:
3sin^2(x)cos^2(x) - sin^4(x) = 0
Factoring out sin^2(x), we get:
sin^2(x)(3cos^2(x) - sin^2(x)) = 0
The first factor sin^2(x) = 0 when x = 0. The second factor 3cos^2(x) - sin^2(x) = 0 when 3cos^2(x) = sin^2(x).
Dividing both sides by cos^2(x), we get:
3 - tan^2(x) = 0
Rearranging, we have:
tan^2(x) = 3
Taking the square root of both sides, we get:
tan(x) = ±√3
The solutions for x are x = π/3 + nπ and x = -π/3 + nπ, where n is an integer.
Next, we need to determine the nature of the critical points.
We can use the second derivative test to determine whether the critical points are minima or maxima. Taking the second derivative of f(x), we get:
f''(x) = 6sin(x)cos^3(x) - 4sin^3(x)cos(x)
Substituting the critical points x = π/3 + nπ and x = -π/3 + nπ into f''(x), we can determine the nature of the critical points.
For x = π/3 + nπ:
f''(π/3 + nπ) = 6sin(π/3 + nπ)cos^3(π/3 + nπ) - 4sin^3(π/3 + nπ)cos(π/3 + nπ)
= 6(√3/2)cos^3(π/3 + nπ) - 4(sin(π/3 + nπ))^3(cos(π/3 + nπ))
= 6(√3/2)(cos^3(π/3 + nπ) - sin^3(π/3 + nπ))
For x = -π/3 + nπ:
f''(-π/3 + nπ) = 6sin(-π/3 + nπ)cos^3(-π/3 + nπ) - 4sin^3(-π/3 + nπ)cos(-π/3 + nπ)
= 6(-√3/2)cos^3(-π/3 + nπ) - 4(sin(-π/3 + nπ))^3(cos(-π/3 + nπ))
= -6(√3/2)(cos^3(-π/3 + nπ) - sin^3(-π/3 + nπ))
Since cos^3(π/3 + nπ) - sin^3(π/3 + nπ) > 0 and cos^3(-π/3 + nπ) - sin^3(-π/3 + nπ) > 0 for all n, we can conclude that the critical points x = π/3 + nπ and x = -π/3 + nπ are minima for the function f(x) = sin^3(x)cos(x).
Therefore, the point of minima of the function sin^3(x)cos(x) are x = π/3 + nπ and x = -π/3 + nπ for all integers n.
The point of minima of function sin^3xcosx is
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