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
As written, it appears that
g(x) = 1+cos(pi/2 - x) = 1+sin(x)...
At x=2, that would make g(2) = 1+sin(2)
since g(2) is not 0, g is not continuous at x=2, since g(x)=0 for x>2.