Question

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
I understand that I need to make sure that it is continuous at x=0 but then what?

Answer 1

\[f(x)=\begin{cases}2-2x-x^2&\text{for }x<0\\2x+2&\text{for }x\ge0\end{cases}\]
For the derivative to exist at \(x=0\), \(f\) must be continuous and be defined at \(x=0\).

You also must show that the derivative from both sides exist are equivalent.

Mathematically, the first condition is
\[\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)~~~~~\text{and}~~~~~f(0)~\text{exists}\]
The second is similar, but you replace \(f\) with \(f'\):
\[\lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)\]
If these limits exist are equal, then \(f'(0)\) is equivalent to these limits.

Answer 2

Differentiating, you have
\[f'(x)=\begin{cases}-2-2x&\text{for }x<0\\?&\text{for }x=0\\2&\text{for }x>0\end{cases}\]
We don't know if the derivative exists at 0 yet, so I'm using the ? as a placeholder.

Answer 3

How did you get -2-2x?

Answer 4

That's the derivative of the parabolic piece, for \(x<0\).

Answer 5

so how do you find out if the derivative exists?

Answer 6

Take the limit from both sides of \(f'(x)\). It's like the procedure for showing continuity.

Answer 7

so as x approaches 0 from the negative f(x)=-2 but as x approaches 0 from the positive f(x)=2 so the derivative does not exist?

Answer 8

Yes, it doesn't exist, so no tangent line exists at 0.

Answer 9

Ok thanks!

Answer 10

You're welcome