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Looking for a and b to ensure continuous function

  • Thread starter ful
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  • #1
ful
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Homework Statement


Find values for a and b that ensure f is a continuous function if

f(x) = ax + 2b if x ≤ 0
x2 +3a - b if 0 < x ≤ 1
2x - 5 if x > 1

Homework Equations





The Attempt at a Solution



ax + 2b = 2x -5 (when x = o)
2b = -5
b = -5/2

3a +5/2 = -3
a = -11/6

when i plug these a and b into the each equation i get
-5 as the answer for the first equation
-3 as the answer for the second equation
-5 as the answer for the third equation

why isnt the 2nd equation coming out to -5? i need them all to equal so they can be continuous. all the limits need to equal.
 

Answers and Replies

  • #2
33,496
5,188

Homework Statement


Find values for a and b that ensure f is a continuous function if

f(x) = ax + 2b if x ≤ 0
x2 +3a - b if 0 < x ≤ 1
2x - 5 if x > 1

Homework Equations





The Attempt at a Solution



ax + 2b = 2x -5 (when x = o)
2b = -5
b = -5/2
No.
When x = 0, you want the first two formulas to produce the same function value.

IOW, when x = 0, you want ax + 2b = x2 + 3a - b

When x = 1, you want the 2nd and 3rd formulas to produce the same function value.


3a +5/2 = -3
a = -11/6

when i plug these a and b into the each equation i get
-5 as the answer for the first equation
-3 as the answer for the second equation
-5 as the answer for the third equation

why isnt the 2nd equation coming out to -5? i need them all to equal so they can be continuous.
Who are "them" and "they"? Do "them" and "they" refer to the same thing?

Try to be more specific here but not using pronouns such is "it" and "they" unless what they refer to is crystal clear.
all the limits need to equal.
No, not at all. You want the left- and right-side limits at x = 0 to be equal, so you work with the 1st and 2nd formulas. You want the left- and right-side limits at x = 1 to be equal, so you work with the 2nd and 3rd formulas.
 
Last edited:
  • #3
ful
6
0
w hen x= 0
ax +2b = 3a -b
b= 3a/2 - b/2

when x=1
x2 +3a -b = 2x-5
1 + 3a-b = 2x-5
3a-b = -4
3a - (3a/2+b/2) = -4
6a +2b = -8

a= -1, b= -1

when I plug in a=-1 b=-1 and x=o to
(ax + 2b) = -2
(x2 +3a -b) = -2

both equations at x=0 give me -2 .is this correct?

the last equation (2x-5) still equals -3
 
  • #4
33,496
5,188
w hen x= 0
ax +2b = 3a -b
b= 3a/2 - b/2
You're really making things hard for yourself.
When x = 0, your first equation should be
2b = 3a - b.

Now move all of the b terms to one side, and then solve for b.
when x=1
x2 +3a -b = 2x-5
When you substitute 1 for x, your equation shouldn't have x in it any more.
1 + 3a-b = 2x-5
3a-b = -4
3a - (3a/2+b/2) = -4
6a +2b = -8

a= -1, b= -1
These aren't the right values. They will give you the same value at x = 0 for the 1st and 2nd formulas, but they give you different values at x = 1 for the 2nd and 3rd formulas. To be continuous, the function has to have the same values at x = 0 (1st & 2nd formulas) and at x = 1 (2nd & 3rd formulas).
when I plug in a=-1 b=-1 and x=o to
(ax + 2b) = -2
(x2 +3a -b) = -2

both equations at x=0 give me -2 .is this correct?


the last equation (2x-5) still equals -3
But you also need to get -3 using the 2nd formula.
 
  • #5
ful
6
0
ok im am trying it again
 
Last edited:
  • #6
ful
6
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i got a= -2 and b=-2
 
  • #7
33,496
5,188
The three separate pieces of this function are all continuous inside their respective intervals, so it's just a matter of figuring out what a and b need to be so that the function is continuous at the two places the three pieces meet: at x = 0 and x = 1.

The values you find for a and b have to make the function continuous at both places.
 
  • #8
ful
6
0
The three separate pieces of this function are all continuous inside their respective intervals, so it's just a matter of figuring out what a and b need to be so that the function is continuous at the two places the three pieces meet: at x = 0 and x = 1.

The values you find for a and b have to make the function continuous at both places.
I understand the concept now. I got a = -2 and b = -2. In general, for a function to be continuous, doesnt the limit and functional value have to equal?
 
Last edited:
  • #9
33,496
5,188
Those are the values. And yes, for a function to be continuous at some point, the limit and function value have to be equal.
 
  • #10
ful
6
0
thank you very much
 

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