anonymous
  • anonymous
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?
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[f(x)=\begin{cases}2-2x-x^2&\text{for }x<0\\2x+2&\text{for }x\ge0\end{cases}\] 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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
How did you get -2-2x?

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anonymous
  • anonymous
That's the derivative of the parabolic piece, for \(x<0\).
anonymous
  • anonymous
so how do you find out if the derivative exists?
anonymous
  • anonymous
Take the limit from both sides of \(f'(x)\). It's like the procedure for showing continuity.
anonymous
  • anonymous
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?
anonymous
  • anonymous
Yes, it doesn't exist, so no tangent line exists at 0.
anonymous
  • anonymous
Ok thanks!
anonymous
  • anonymous
You're welcome

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