anonymous
  • anonymous
Please help with math!!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Need an explanation on how to do this.
anonymous
  • anonymous
@phi

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@mathmath333
anonymous
  • anonymous
@freckles
anonymous
  • anonymous
@imqwerty
anonymous
  • anonymous
@triciaal
Owlcoffee
  • Owlcoffee
We define continuity for a function "f" as the coicident lateral limit of the intermediate point. We can traduce like this: \[\lim_{x \rightarrow a ^{\pm}} f(x)=f(a)\] What does this mean? In very simple words, the sufficient and necessary condition for a function to be continous at a point as long as the lateral limits of such point have a value coincident with the image of that point. Let's take a look at that and apply it to the problem in question, as you can see. Since the function is divided we will study what happens on that x=-4 point. So we will study the two following limits: \[\lim_{x \rightarrow -4^{-}}\frac{ x+4 }{ x^2-16 }\] and \[\lim_{x \rightarrow -4^{+}}\frac{ c }{ x+12 }\] These two limits arise from the very definition of limits, since when \(x<-4\), the function f takes the structure of \(\frac{ x+4 }{ x^2-16 }\) and when \(x \ge -4\) the function f takes the structure \(\frac{ c }{ x+12 }\). Now, solving those limits will give a value, and then you will want to use "c" as a variable to make them equal, therefore making the function continous.
anonymous
  • anonymous
But when you plug in -4 they equal 0 so how can you equal both equations together?
anonymous
  • anonymous
@Owlcoffee
Owlcoffee
  • Owlcoffee
Let's first work with: \[\lim_{x \rightarrow (-4)^-} \frac{ x+4 }{ x^2-16 }=\frac{ 0 }{ 0 }\] This is an indetermination, which means that that you'll have to apply some factorization strategies in order to lift up the indetermination and solve the limit.
anonymous
  • anonymous
So then it would be \[\frac{ 1 }{ (x-4) }\] right?
anonymous
  • anonymous
@Owlcoffee
Owlcoffee
  • Owlcoffee
Yes, correct, and then you plug "-4", what do you get?
anonymous
  • anonymous
-1/8
anonymous
  • anonymous
Then what do I do?
Owlcoffee
  • Owlcoffee
Good, now solve the second limit: \[\lim_{x \rightarrow (-4)^+}\frac{ c }{ x+12 }\] In this case, just treat "c" as a constant and apply the limit.
anonymous
  • anonymous
\[\frac{ -1 }{ 8 }=\frac{ c }{ 8 }\]\[-8=8c\]\[-1=c\]
anonymous
  • anonymous
Is that right?
Owlcoffee
  • Owlcoffee
It is correct.
anonymous
  • anonymous
Okay thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.