anonymous
  • anonymous
f(x)= -x^2+2, 0≤ x <1 2, x=1 x^2, 1
Calculus1
  • 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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
jim_thompson5910
  • jim_thompson5910
how far did you get?
anonymous
  • anonymous
i think a) is 4 but I am not sure

Looking for something else?

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

More answers

jim_thompson5910
  • jim_thompson5910
a) is 4, nice work
jim_thompson5910
  • jim_thompson5910
you start on the left side of x = 2, then you slowly approach x = 2 along the graph and you'll get closer and closer to y = 4
jim_thompson5910
  • jim_thompson5910
how about b?
anonymous
  • anonymous
Idk how to do b :/
jim_thompson5910
  • jim_thompson5910
what is the limiting value of f(x) as x ---> 1
jim_thompson5910
  • jim_thompson5910
ie if you approach x = 1 from both directions, what does y get closer and closer to?
anonymous
  • anonymous
2?
jim_thompson5910
  • jim_thompson5910
no you approach x = 1 this means you get closer and closer to x = 1...but you actually never officially arrive at x = 1
jim_thompson5910
  • jim_thompson5910
does that make sense?
anonymous
  • anonymous
yes
anonymous
  • anonymous
what about f(1)?
jim_thompson5910
  • jim_thompson5910
f(1) is the y value of the point at x = 1
jim_thompson5910
  • jim_thompson5910
the limit is slowly approaching x = 1 (it never gets there officially, just gets closer and closer) f(1) is actually at x = 1 (it's officially there and stopped)
anonymous
  • anonymous
so f(1) has a greater value
jim_thompson5910
  • jim_thompson5910
good, the limiting value as x ----> 1 is 1 but f(1) = 2 so you are correct
anonymous
  • anonymous
idk how to do c:/
jim_thompson5910
  • jim_thompson5910
what are some points where the limit does not exist?
anonymous
  • anonymous
where the holes are at, i think
jim_thompson5910
  • jim_thompson5910
that's a good guess, but look how the limit exists at x = 1 this is because you can approach x = 1 from either side and arrive at the same y value
jim_thompson5910
  • jim_thompson5910
however, the limit does NOT exist at x = 2, why is this?
anonymous
  • anonymous
cause the lim f(x) as x approaches 2- is 4 and the lim f(x) as x approaches 2+ is 3 so limf(x) as x approaches 2 does not exist because those two limits aren't equal
jim_thompson5910
  • jim_thompson5910
bingo
jim_thompson5910
  • jim_thompson5910
the left and right hand limits are NOT equal, so the limit does NOT exist
jim_thompson5910
  • jim_thompson5910
what's another point where the limit doesn't exist?
anonymous
  • anonymous
I am not sure :/
jim_thompson5910
  • jim_thompson5910
hint: it's not in the middle
anonymous
  • anonymous
wait so the limit f(x) as x approaches 2, does not exist, is part of my answer?
jim_thompson5910
  • jim_thompson5910
yes, there's another point though
jim_thompson5910
  • jim_thompson5910
c = 2 is one value but there is one more value of c
anonymous
  • anonymous
is it c=0
jim_thompson5910
  • jim_thompson5910
yep, the limit does NOT exist at x = 0
jim_thompson5910
  • jim_thompson5910
because the left hand limit does not exist
jim_thompson5910
  • jim_thompson5910
both the left and right hand limits must exist and they must be equal for the overall limit to exist
anonymous
  • anonymous
why doesn't the left hand limit exist?
jim_thompson5910
  • jim_thompson5910
because the function ends at x = 0, so there is no bit to the left to start approaching from
jim_thompson5910
  • jim_thompson5910
the general rule is that the limit does not exist at endpoints because you can't approach from beyond the function
anonymous
  • anonymous
ooh okay got it:)
jim_thompson5910
  • jim_thompson5910
that's great
anonymous
  • anonymous
thank you:)
jim_thompson5910
  • jim_thompson5910
you're welcome

Looking for something else?

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