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How to make a function continuous

  • Thread starter ammsa
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  • #1
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Homework Statement


consider the function f(x) = (4 - x) / (2 - [tex]\sqrt{x}[/tex]). define a new function g(x) = f(x) for all x except 4 and such g(x) is continuous at 4.


The Attempt at a Solution



i got the limit of the f(x) when x approches 4 and i got 4 as the final answer.
here's how i did it,

1) (4-x) / (2 - [tex]\sqrt{x}[/tex] )
2) (2 + [tex]\sqrt{x}[/tex]) (2 - [tex]\sqrt{x}[/tex]) / (2 - [tex]\sqrt{x}[/tex])
3) (2 + [tex]\sqrt{x}[/tex])
4) when we plug in 4 in the equation (2 + [tex]\sqrt{x}[/tex]), we get
(2 + [tex]\sqrt{4}[/tex]) = 4

so, g(x) = f(x) when x [tex]\neq[/tex] 4
and = (2 + [tex]\sqrt{x}[/tex]) when x = 4



now my question is is my way of approching this problem correct, and is my answer correct?
 
Last edited:

Answers and Replies

  • #2
Char. Limit
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I'd say your way of approaching the problem is correct, since it got the correct answer.
 
  • #3
tiny-tim
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hi ammsa! :smile:

(have a square-root: √ :wink:)

yup, your steps 1) to 4) are perfect :approve:

(but in the last line it would be simpler to say "g(x) = 4 when x = 4")
 
  • #4
HallsofIvy
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By the way, we would say that the function
[tex]f(x)= \frac{x- 4}{x- \sqrt{2}}[/tex]
has a 'removable' discontinuity at [itex]x= \sqrt{2}[/itex]. That is why it was possible to make it continuous.

It is also correct that the [itex]g(x)= x+ \sqrt{2}[/itex] is NOT the same function as f.

Some people make the mistake of writing
[tex]\frac{x-4}{x-\sqrt{2}}= \frac{(x-\sqrt{2})(x+\sqrt{2})}{x- \sqrt{2}}= x+ \sqrt{2}[/tex]
but that last inequality is only true for [itex]x\ne \sqrt{2}[/itex]and not for all x.
 

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