Find the critical numbers of each given function.
(a) f(x)=3rt(x^(2)-x
(b) g(x)=2sinx+2cos^(2)x, 0 <=x<=2pie
To find the critical numbers of a function, we need to find the values of x where the derivative of the function is either zero or undefined.
(a) f(x) = 3√(x^2 - x)
To find the derivative, we can use the power rule for differentiation and the chain rule:
f'(x) = (1/2)(x^2 - x)^(-1/2)(2x - 1)
= (x - 1)/(√(x^2 - x))
Now, to find the critical numbers, we need to set the derivative equal to zero and solve for x:
(x - 1)/(√(x^2 - x)) = 0
For this equation to be true, the numerator (x - 1) must be zero:
x - 1 = 0
x = 1
Therefore, the critical number for the function f(x) = 3√(x^2 - x) is x = 1.
(b) g(x) = 2sin(x) + 2cos^2(x), 0 <= x <= 2π
To find the derivative, we differentiate each term separately:
g'(x) = 2cos(x) - 4sin(x)cos(x)
Next, we set the derivative equal to zero and solve for x:
2cos(x) - 4sin(x)cos(x) = 0
Factoring out the common factor of 2cos(x):
2cos(x)(1 - 2sin(x)) = 0
Setting each factor equal to zero:
2cos(x) = 0 or 1 - 2sin(x) = 0
For the first equation, 2cos(x) = 0, cos(x) must be zero:
cos(x) = 0
x = π/2 or 3π/2
For the second equation, 1 - 2sin(x) = 0, we solve for sin(x):
-2sin(x) = -1
sin(x) = 1/2
This occurs at x = π/6 or x = 5π/6.
Therefore, the critical numbers for the function g(x) = 2sin(x) + 2cos^2(x), in the interval 0 <= x <= 2π, are x = π/2, 3π/2, π/6, and 5π/6.