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?

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

To determine whether the graph has a tangent at x=0, we need to check if the function is continuous at this point. In order for a function to have a tangent at a certain point, it should be continuous at that point.

In this case, we have a piecewise function with two separate equations:

1) f(x) = 2 - 2x - x^2, for x < 0
2) f(x) = 2x + 2, for x ≥ 0

To check if the function is continuous at x=0, we need to compute the limit of the function as x approaches 0 from both sides and see if they match.

For the left-hand side (x<0):
lim(x→0-) f(x) = lim(x→0-) (2 - 2x - x^2)
= 2

For the right-hand side (x≥0):
lim(x→0+) f(x) = lim(x→0+) (2x + 2)
= 2

Since both limits give us the same value of 2, we can conclude that the function is continuous at x=0. Therefore, it does have a tangent at this point.

The reason why the answer key might say that it does not exist could be a mistake or a misinterpretation. However, based on the calculations, we can confidently say that the graph does have a tangent at x=0.