Question

Find the critical points of f(x)=2sinx+2cosx and determine the extreme values on [0,2π]. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)

critical points x= ___________x
maximum f(x)=__________-x
minimum f(x)=___________x

Answer 1

The critical points of f(x)=2sinx+2cosx on [0,2π] are x=arcsin((2-√2)/2), and the extreme values are minimum f(x)≈-0.828 and maximum f(x)=2.

Step-by-step explanation:

To find the critical points of f(x) = 2sin(x) + 2cos(x) on the interval [0,2π], we need to take the derivative and set it equal to zero. Let's find the derivative of f(x):

f'(x) = 2cos(x) - 2sin(x)

Now, let's set f'(x) = 0 and solve for x:

2cos(x) - 2sin(x) = 0

cos(x) - sin(x) = 0

(1 - √2/2)sin(x) = cos(x)

sin(x) = (2 - √2)/2

x = arcsin((2 - √2)/2)

So the critical point is x = arcsin((2 - √2)/2) ≈ 0.785

To determine the extreme values on [0,2π], we need to evaluate f(x) at the critical points and the endpoints of the interval. Let's plug in the values:

f(0) = 2sin(0) + 2cos(0) = 2(0) + 2(1) = 2

f(2π) = 2sin(2π) + 2cos(2π) = 2(0) + 2(1) = 2

f(arcsin((2 - √2)/2)) ≈ -0.828

So the minimum value of f(x) on [0,2π] is approximately -0.828 and the maximum value is 2.