Question

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?

Answer 1

\[f(x)=\begin{cases}x-4&\text{for }1\le x<2\\\\\\\frac{1}{x-3}&\text{for }2\le x<5\\\\\\
\frac{11}{2}-x&\text{for }5\le x\end{cases}\]
Is this the function?

Answer 2

everything is the same except for the last one it is -x+5.5

Answer 3

Right, well 5.5 is 11/2.

Answer 4

oh yea lol

Answer 5

Anyway, to show \(f(x)\) is continuous at \(x=5\), you have to show that
\[\lim_{x\to5^-}f(x)=\lim_{x\to5^+}f(x)\]

Answer 6

oh then how do I show that?

Answer 7

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
And that \(f(5)\) exists and is equal to these limits, but that's given by the function.
\(\color{blue}{\text{End of Quote}}\)

Answer 8

\[\lim_{x\to5^-}\frac{1}{x-3}=\lim_{x\to5^+}\left(\frac{11}{2}-x\right)\]
Is this true?

Answer 9

yes!

Answer 10

Okay, so \(f\) is indeed continuous at 5.

Answer 11

For the second part, think about what values in the desired interval make the function undefined.

Answer 12

but what does it mean by interval [1,4]?