anonymous
  • anonymous
Determine whether f(x) approaches negative or positive infinity as x approaches 4 from the left and right. f(x)= 1/(x-4)^2
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[f(x)=\frac{ 1 }{ x-4^{2} }\]
hartnn
  • hartnn
you know what "x approaches 4 from left" mean?
anonymous
  • anonymous
In the negative direction, correct?

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hartnn
  • hartnn
yes, more precisely, it means value of x is very very very near to 4 but LESS than 4 like 3.9 or 3.9999999
anonymous
  • anonymous
Okay, and from there I can come to the conclusion that it will be infinitely positive and infinitely negative as x approaches 4 from the left and right?
hartnn
  • hartnn
woah! slow down.. there are 2 cases there when x approaches 4 from left, then is x-4 negative or positive?? is (x-4)^2 negative or positive?? based on that you will be able to tell, whether its +infinity or -infinity
hartnn
  • hartnn
similarly, the other case when x approaches 4 from right, then is x-4 negative or positive?? is (x-4)^2 negative or positive?? based on that you will be able to tell, whether its +infinity or -infinity
anonymous
  • anonymous
from the right is positive infinity and from the left is negative infinity?
hartnn
  • hartnn
can you first answer these?? is x-4 negative or positive?? is (x-4)^2 negative or positive??
anonymous
  • anonymous
using what values? 3.999?
hartnn
  • hartnn
if you're consider x is approaching from left, then yes
anonymous
  • anonymous
then x-4 would be negative (x-4)^2 is postive
hartnn
  • hartnn
good! so for x approaching from left we get positive infinity :)
hartnn
  • hartnn
now take x approaching from right case
anonymous
  • anonymous
then it would be negative infinity
anonymous
  • anonymous
I understand now, sorry for not understanding it until now and thank you for helping me!
hartnn
  • hartnn
welcome ^_^
hartnn
  • hartnn
but actually for right case also, it will be +infinity
hartnn
  • hartnn
because (x-4)^2 will anyways be positive
hartnn
  • hartnn
@Alyssa8
anonymous
  • anonymous
im not that smart lol sorry

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