for what value of a is the function continuous at every x?

[deuxillusion]

continuity question. should be easy, just need help going in the right direction.

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'm thrown off here a little. 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?

 

[daon2]

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

can someone explain the steps for finding the answer? i have no idea how to solve this problem. i missed class, so i've taught myself the basic concepts of continuity. graphically i think i understand but i'm not getting it algebraically. i don't even know what this question is asking.

this is what i understand so far... To find out if a function is continuous at point a, you have to find the limit and the value of a when plugged into the function. Then if these two equal each other, the function is continuous.

You seem to be a little confused. You want to find all value(s) of 'a' for which the left and right limits are equal at -1, and such that the function's value at -1 is equal to that limit.

the limit as x approaches -1 from the right would be 0 (correct?)

Check again. Plug in -1, and you get 1+a.

 

[deuxillusion]

oh okay, i wasn't even doing limits right, lol.

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'm thrown off here a little. 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?

 

[deuxillusion]

i'm sorry if i'm being obnoxious by posting again. i have a quiz in an hour and i still don't understand this, so if anyone is on who can get me going in the right direction, that would be awesome in the mean time i'll try to be patient and work on it myself..

 

[daon2]

oh okay, i wasn't even doing limits right, lol.

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.

You don't know anything about a, so how do you know they aren't equal? If there was no a that made those two quantities equal, then you may conclude the function is not continuous at -1. Again the question asks you to find a (if possible) so f(x) is continuous. Setting them equal and solving gives a=2. So the limit will exist, and has value 3, if a = 2. Then it just so happens that f(-1)=3 also when a=2. So f(x) is continuous at -1 for this value a.

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?

I just read this part now, sorry. So your intuition was correct.

 

[pka]

i'm sorry if i'm being obnoxious by posting again. i have a quiz in an hour and i still don't understand this, so if anyone is on who can get me going in the right direction, that would be awesome in the mean time i'll try to be patient and work on it myself..

You make \(\displaystyle \displaystyle\lim _{x \to -1^ - } f(x) = \lim _{x \to -1^ + } f(x)\)