Question

graph is a piecewise; f(x)= 1) 2-2x-x^2, x<0 and 2) 2x+2, x>=0. Determine whether graph has a tangent at x=o. If it does not, explain why not.

I have no idea. When I plug in x=0, both equations give 2, but answer key says it does not exist. Explain?

Answer 1

f1(0) = 2
f2(0) = 2
So, f is continuous. If a tangent exists, the slope from the left must be the same as the slope from the right.

f1'(x) = -2-2x
f2'(x) = 2

f1'(0) = -2

So, there is no tangent. You can see this from the graphs below. They touch at (0,2), but the slope changes abruptly there.

http://www.wolframalpha.com/input/?i=plot+y%3D2-2x-x^2%2C+y%3D2x%2B2