anonymous
  • anonymous
Can I please have some help solving this piece-wise function/limit?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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misty1212
  • misty1212
sure
misty1212
  • misty1212
they are usually not nearly as hard as they look
anonymous
  • anonymous

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misty1212
  • misty1212
do we really have to sketch them?
misty1212
  • misty1212
or can we just find the limits if they exist? that is much easier than drawing
anonymous
  • anonymous
I'm sure they aren't eventually, I just need more practice with them. I have to sketch them, if you just show me how to go about finding the limit, I can draw it afterward.
misty1212
  • misty1212
ok lets start with the first one
anonymous
  • anonymous
Thank you!
misty1212
  • misty1212
the definition of the function changes at 2 right?
anonymous
  • anonymous
Yes
misty1212
  • misty1212
so to see if it is continuous at 2, all you need to do is plug in 2 for both definitions of \(f\) in other words compute \[2^2\] and \(8-2\times 2\)
misty1212
  • misty1212
i get \(4\) both times, i bet you do too that means the limit exists, since it is the same in both definitions, and \[\lim_{x\to 2}f(x)=4\]
misty1212
  • misty1212
guess what we do next?
anonymous
  • anonymous
So I can just input all values of two for x? Excluding the third. What exactly makes this limit a piece-wise function?
misty1212
  • misty1212
we are not done yet, we just checked the limit as x goes to 2
misty1212
  • misty1212
and yah just plug it in
anonymous
  • anonymous
we also need to check if it goes to 4?
misty1212
  • misty1212
right
anonymous
  • anonymous
plug in 4 as well?
misty1212
  • misty1212
because only at the change in definition can something go wrong otherwise it is a bunch of polynomials which are always continuous
misty1212
  • misty1212
yeah plug in 4 to the middle one and the bottom one and see if you get the same answer (you do not)
anonymous
  • anonymous
Okay, thank you for explaining.
anonymous
  • anonymous
Since I have to plug 4 into the functions
misty1212
  • misty1212
you are quite welcome told you it was easy right? math teachers often make it seem harder than it is
misty1212
  • misty1212
yeah but only in the middle one and the bottom one where it changes from \[8-2x\] to \[4\]
anonymous
  • anonymous
Since i have to plug 4 into the functions, does that mean that 4 is not part of the limit?
misty1212
  • misty1212
i am not sure what you are asking
misty1212
  • misty1212
if you plug 4 in to \[8-2x\] you get 0 right?
anonymous
  • anonymous
that the limit for the function exists everywhere else but 4. is that the correct?
anonymous
  • anonymous
yes, that's what i got when i plugged it in.
misty1212
  • misty1212
ok and to the right,the function is a constant, it is just 4
misty1212
  • misty1212
so that means, since \(0\neq 4\) you have \[\lim_{x\to 4}f(x)\] does not exist
misty1212
  • misty1212
give me a second and i will show you a graph
misty1212
  • misty1212
taking me a second, but soon
anonymous
  • anonymous
so my final answer would be lim f(x) x->2 and lim f(x) x-4 does not exist
anonymous
  • anonymous
you're fine, take your time. i really appreciate the help.
misty1212
  • misty1212
yes picture almost done damn syntax is killing me
misty1212
  • misty1212
copy and paste http://www.wolframalpha.com/input/?i=Plot%5BPiecewise%5B%7B%7Bx%5E2%2C+x+%3C+2%7D%2C+%7B8-2x+%2C+2%3Cx+%3C4%7D%2C%7B4%2Cx%3E4%7D%7D%5D%2C+%7Bx%2C+0%2C+6%7D%5D
misty1212
  • misty1212
oh just click on it
misty1212
  • misty1212
you can see that it is continuous at 2, there is no break there but there is a jump from 0 to 4 at 4, so not continuous there
misty1212
  • misty1212
and yes, that is your final answer for the limits from the first question graph is what i linked to
anonymous
  • anonymous
okay, thank you so much! that was incredibly helpful! I'll try the next one on my own. :)
misty1212
  • misty1212
here is the second one easier since i had the template http://www.wolframalpha.com/input/?i=Plot%5BPiecewise%5B%7B%7Bsin%28x%29%2C+x+%3C+0%7D%2C+%7B1-cos%28x%29+%2C+0%3Cx+%3Cpi%7D%2C%7Bcos%28x%29%2Cx%3Epi%7D%7D%5D%2C+%7Bx%2C+-2%2C+6%7D%5D
misty1212
  • misty1212
oh if you want to try it on your own, don't click, although the picture may be helpful
misty1212
  • misty1212
\[\color\magenta\heartsuit\]
anonymous
  • anonymous
@misty1212 very good explanations!

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