g(x)={1+cos(pi/2-x/1)
0
for-2 <or= x <or=2
for x<2 or x>2
is g(x) continuous at x=2 is it differentiable at x=2? ex[lain in detail I'm very confused as to how to solve this problem.
To determine if the function g(x) is continuous at x = 2, we need to check if the left and right limits of g(x) at x = 2 exist and are equal to the value of g at x = 2.
1. Calculate the limit from the left side (x approaches 2 from the smaller side):
Lim(x→2-) g(x) = Lim(x→2-) (1 + cos(π/2 - x/1))
Substituting x = 2:
Lim(x→2-) g(x) = Lim(x→2-) (1 + cos(π/2 - 2/1))
= Lim(x→2-) (1 + cos(π/2 - 2))
= Lim(x→2-) (1 + cos(-π/2))
= Lim(x→2-) (1 + cos(-90°))
= Lim(x→2-) (1 + cos(-π/2))
= Lim(x→2-) (1 + 0)
= 1
2. Calculate the limit from the right side (x approaches 2 from the larger side):
Lim(x→2+) g(x) = Lim(x→2+) 0
As x approaches 2 from the right side, the function becomes constant at 0.
3. Calculate the value of g(x) at x = 2:
g(2) = 0
Now, to determine if the function is continuous at x = 2, we compare the left and right limits with the value of g(2):
Lim(x→2-) g(x) = 1
Lim(x→2+) g(x) = 0
g(2) = 0
Since the left and right limits are equal to the value of g(2), the function g(x) is continuous at x = 2.
To determine if the function g(x) is differentiable at x = 2, we need to check if the derivative exists at that point.
Since g(x) is not continuous at x = 2 (as we have already determined), it cannot be differentiable at that point. Differentiability implies continuity, meaning if a function is differentiable at a point, it must be continuous at that point as well. Therefore, g(x) is not differentiable at x = 2.