Continuity

Alexander Lorien

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Is the statement, "If ๐‘“ is a continuous function at a, then so is |๐‘“|." true or false? If true, give a formal proof. If false, provide a counterexample.
 

Subhotosh Khan

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Is the statement, "If ๐‘“ is a continuous function at a, then so is |๐‘“|." true or false? If true, give a formal proof. If false, provide a counterexample.
What is the formal definition of a continuous function?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 

pka

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Is the statement, "If ๐‘“ is a continuous function at a, then so is |๐‘“|." true or false? If true, give a formal proof. If false, provide a counterexample.
Do you know that \(||a|-|b||\le|a-b|~?\)
\(|a|=|a-b+b|\le|a-b|+|b|\) implies \(|a|-|b|\le|a-b|\)
Likewise \(|b|-|a|\le|b-a|=|a-b|\) or \(-|a-b|\le||a|-|b||\le|a-b|\)
So \(\large\left||a|-|b|\right|\le|a-b|\)
 
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Jomo

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Can you draw some continuous functions for f(x) and see if |f(x)| is continuous? Make sure that your f(x) takes on both positive and negative values.

Post back with your results including pictures of the functions.
 

Alexander Lorien

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Does this mean that the statement is false??
 

Alexander Lorien

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Do you know that \(||a|-|b||\le|a-b|~?\)
\(|a|=|a-b+b|\le|a-b|+|b|\) implies \(|a|-|b|\le|a-b|\)
Likewise \(|b|-|a|\le|b-a|=|a-b|\) or \(-|a-b|\le||a|-|b||\le|a-b|\)
So \(\large\left||a|-|b|\right|\le|a-b|\)
Does it mean that the statement is false? Sorry i am confused
 

pka

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Does it mean that the statement is false? Sorry i am confused
You simply need to sit down with an instructor. You need help.
If \(f\) is a contusions at \(x=a\) then if \(c>0\) then \(\exists d>0\) so that \(|x-a|<c\to |f(x)-f(a)|<d\)
BUT \(||f(x)|-|f(a)||\le|f(x)-f(a)| <d\) Does that mean that mean that \(|f|\) is continuous at \(x=a~?\)
 

Jomo

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Does this mean that the statement is false??
Draw a few functions and see for yourself. You need to see if it is true or not. A few simple drawing will do wonders.
 
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