Let g(x)=|cos x-1|. The maximum value attained by g on the closed interval [0, 2pi] is for x equal to
A) -1
B) 0
C) pi/2
D) 2
E) pi
I think its pi/2 but I'm not sure at all.
I think it's 0.
Look at the graph. There's plenty of graphers online.
But since the interval of the normal cos function is 1, -1, cos(x)-1 is 0, -1
which means 0 is the biggest value
look... im not sure as well about which one it is, but im sure that its not C, D or E.... cause just like the guy above said, the interval of the normal cos function will always be 1,-1... so u wouldnt be able to get those values.... the maximum would be 0 and the lowest would be -2...so yea... its most probably 0
To find the maximum value of g(x) on the closed interval [0, 2π], we need to evaluate g(x) at the critical points and endpoints of the interval.
The function g(x) = |cos x - 1| can be split into two cases based on the sign of (cos x - 1):
Case 1: cos x - 1 ≥ 0
In this case, |cos x - 1| = cos x - 1.
Case 2: cos x - 1 < 0
In this case, |cos x - 1| = -(cos x - 1) = 1 - cos x.
Now let's find the critical points:
Case 1: cos x - 1 ≥ 0
For cos x - 1 = 0,
cos x = 1
The critical point occurs at x = 0.
Case 2: cos x - 1 < 0
For cos x - 1 = 0,
cos x = 1
No critical point occurs in this case.
Now let's evaluate g(x) at the endpoints of the interval:
g(0) = |cos 0 - 1| = 1 - 1 = 0
g(2π) = |cos (2π) - 1| = |1 - 1| = 0
Thus, the maximum value of g(x) is 0, and it is attained at the critical point x = 0 and the endpoints x = 0 and x = 2π.
Therefore, the correct answer is:
B) 0
To find the maximum value of the function g(x) on the closed interval [0, 2pi], we need to evaluate g(x) at the critical points and endpoints of the interval.
First, let's find the critical points by finding where the derivative of g(x) is equal to zero or does not exist. The derivative of g(x) with respect to x can be calculated using the chain rule as follows:
g'(x) = [d(|cos x - 1|) / d(cos x - 1)] * [d(cos x - 1) / dx]
The derivative of |cos x - 1| with respect to (cos x - 1) can be found as:
d(|u|) / du = u / |u|
Using this, we have:
g'(x) = [cos x - 1] / |cos x - 1| * [d(cos x - 1) / dx]
The derivative of (cos x - 1) with respect to x is (-sin x). Therefore:
g'(x) = [(cos x - 1) / |cos x - 1|] * (-sin x)
Next, we set the derivative g'(x) equal to zero:
[(cos x - 1) / |cos x - 1|] * (-sin x) = 0
We have two cases to consider:
Case 1: cos x - 1 = 0
If cos x - 1 = 0, then cos x = 1.
The only solution in the interval [0, 2pi] for cos x = 1 is x = 0.
Case 2: |cos x - 1| = 0
If |cos x - 1| = 0, then cos x - 1 = 0.
Solving for cos x, we have cos x = 1.
The only solution in the interval [0, 2pi] for cos x = 1 is x = 0.
Therefore, the only critical point in the interval [0, 2pi] is x = 0.
Next, we evaluate g(x) at the critical point and endpoints of the interval to find the maximum value:
g(0) = |cos 0 - 1| = |1 - 1| = 0
g(2pi) = |cos(2pi) - 1| = |1 - 1| = 0
Therefore, the maximum value attained by g(x) on the closed interval [0, 2pi] is 0.
Since none of the answer choices include 0, none of the options given are correct.