Calculus problem?

Question

Let f(x) = sin x , for x greater than / equal to 0
= - sin x, for x < 0
Then f(x) is
continuous at x=0

discontinuous at x=0
differentiable at x=0
not differentiable at x=0
Please explain

Answer 1

What does the graph of this function look like?

Starting at x=0 and moving to the right, it is an ordinary sine curve. Starting just below x=0, and moving to the left, it is an inverted sine curve, creating a mirror image about the y-axis. Thus it comes to a point at (0,0), a sharp valley between two delicious breasts. It is continuous.

What about the slopes for x>0 and x<0? Just to the right of x=0, the slope is cos(0), or positive 1. Just to the left of x=0, the slope is -cos(0), or negative 1. Thus it is not differentiable.

Answer 2

Well definitely f(x) is discontinuous.I too faced similar problems when I was at school,anyways you probably might be thinking y f(x) is discontinuous at x=0 when f(x) has a value .But u see u may call a function to be continuous only if u r able to draw a graph of the function at all points(ALL x values).Now tell me a simple answer what is the slope of f(x) when u move from x<0 to x>0 u see there's no correct answer the reason behind is we do not know the value of f(x) when x tends to some positive value and is thus discontinuous.
Note: The function would have been continuous if f(x)= -sinx for x<=0.
Since it is discontinuous it is not differentiable.

Answer 3

Wolverine is incorrect. The function is continuous at x = 0 because the limit from the right equals the limit from the left which is equal to f(0) = 0

He is correct, however, that the function is not differentiable. If you look at the slope of the tangent line approaching from the left, it is not the same as the slope of the tangent line approaching from the right.