Determine whether the curve has a tangent at the indicated point. if it it does, give its slope. if not, explain why not.. f(x)= 2-2x-x^2, x< 0. 2x+2, x > (or equal to) 0. at x = 0

Since the function is piecewise defined, we must see if it's continuous at the given point. Which means that it must have the same right and left limits. In this case we have from the left (x<0): f(0-) = 2-2*0-0^2 = 2 and from the right (x>=0): f(0+) = 2*0 + 2 = 2 They are the same, so the function is continuous there. We also need the function to be smooth at the point for it to have a tangent, which means that the derivative also must be continuous at x=0, we have: f'(x) = -2-2x if x<0 = 2 if x>=0 So f'(0-) = -2-2*0 = -2 and f'(0+) = 2, which is not the same, so it's not smooth. That means that we can't define a tangent line at x=0 (the function has a spike at x=0).

The answer is e &