anonymous
  • anonymous
Define f(x)= x-4, 1≤x<2 1/x-3, 2≤x<5 -x+5.5, 5≤x a) show that f(x) is continuous at x=5 b) where on the interval [1,4] is f discontinuous?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[f(x)=\begin{cases}x-4&\text{for }1\le x<2\\\\\\\frac{1}{x-3}&\text{for }2\le x<5\\\\\\ \frac{11}{2}-x&\text{for }5\le x\end{cases}\] Is this the function?
anonymous
  • anonymous
everything is the same except for the last one it is -x+5.5
anonymous
  • anonymous
Right, well 5.5 is 11/2.

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anonymous
  • anonymous
oh yea lol
anonymous
  • anonymous
Anyway, to show \(f(x)\) is continuous at \(x=5\), you have to show that \[\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)\]
anonymous
  • anonymous
oh then how do I show that?
anonymous
  • anonymous
\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles And that \(f(5)\) exists and is equal to these limits, but that's given by the function. \(\color{blue}{\text{End of Quote}}\)
anonymous
  • anonymous
\[\lim_{x\to5^-}\frac{1}{x-3}=\lim_{x\to5^+}\left(\frac{11}{2}-x\right)\] Is this true?
anonymous
  • anonymous
yes!
anonymous
  • anonymous
Okay, so \(f\) is indeed continuous at 5.
anonymous
  • anonymous
For the second part, think about what values in the desired interval make the function undefined.
anonymous
  • anonymous
but what does it mean by interval [1,4]?

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