deuxillusion
New member
- Joined
- Jan 18, 2012
- Messages
- 7
i'm guessing this should be fairly easy if you know what you're doing, it was a problem in our notes, but i'm not understanding. i have a quiz over it in about an hour though, so any help would be greatly appreciated!
1. For what value of a is the function continuous at every x?
and i get a piecewise function
f(x){x3 - 2ax , x < -1 and 1 - ax , x ≥ -1
to find the limit, i would plug in -1 for x and solve.
The lim as x approaches -1+ = 1 + a
The lim as x approaches -1- = -1 + 2a
and since these aren't equal, the limit as x approaches -1 does not exist.
(i think that's right, let me know if i'm wrong)
this is where i'm thrown off. if the limit doesn't exist, then there has to be some kind of discontinuity in the graph of this function. if there's a discontinuity, then how is it possible to find an a value at which every point is continuous? i know it has to be possible or he wouldn't have given us this question... but it doesn't seem like it should be possible. what am i missing?
my guess is that i would set the two limits equal to each other (1+a = -1+2a) and solve for a. is that close?
1. For what value of a is the function continuous at every x?
and i get a piecewise function
f(x){x3 - 2ax , x < -1 and 1 - ax , x ≥ -1
to find the limit, i would plug in -1 for x and solve.
The lim as x approaches -1+ = 1 + a
The lim as x approaches -1- = -1 + 2a
and since these aren't equal, the limit as x approaches -1 does not exist.
(i think that's right, let me know if i'm wrong)
this is where i'm thrown off. if the limit doesn't exist, then there has to be some kind of discontinuity in the graph of this function. if there's a discontinuity, then how is it possible to find an a value at which every point is continuous? i know it has to be possible or he wouldn't have given us this question... but it doesn't seem like it should be possible. what am i missing?
my guess is that i would set the two limits equal to each other (1+a = -1+2a) and solve for a. is that close?