Continuity - Finding two variables such that a function becomes continuous.

JP_Mtl

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Joined
Jan 31, 2015
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Hey folks,

I'm currently taking Calculus I, and I was recently stumped by the following problem (having to do with making a piecewise function continuous):
Note: I'm taking calculus after about 10 years away from algebra, so this may actually be an algebra problem in disguise.

Problem: Find a and b such that f(x) is continuous.
f(x) =
ax^2+b for x<-1
4x-1 for -1≤x<1
ax^2+bx+8 for x≥1

---
My work:
I started by equating ax^2+b = 4x-1 and substituting x=-1. This gives a+b = -5.
Then, I equated 4x-1 = ax^2+bx+8 and substituting x=1. This also gives a+b = -5.

This is where I ran into difficulties - I think there is no way to solve for a and b using elimination, since both equations generated above are identical, and I tried "isolating" a as well as b from the equation a+b=-5, and plugging this back into one of the equations (ex. 4x-1=ax^2+(-5-a)+8 and substitue x=1), but this only generates tautologies such as 3=3.

I also tried plugging different values for a and b into a graphing program, to no avail (and ideally I'd still need to find an algebraic proof too.)

Is there something I'm missing here? Is there possibly no single solution to this problem? Or, another way to solve for a and b?
 
Hey folks,

I'm currently taking Calculus I, and I was recently stumped by the following problem (having to do with making a piecewise function continuous):
Note: I'm taking calculus after about 10 years away from algebra, so this may actually be an algebra problem in disguise.

Problem: Find a and b such that f(x) is continuous.
f(x) =
ax^2+b for x<-1
4x-1 for -1≤x<1
ax^2+bx+8 for x≥1

---
My work:
I started by equating ax^2+b = 4x-1 and substituting x=-1. This gives a+b = -5.
Then, I equated 4x-1 = ax^2+bx+8 and substituting x=1. This also gives a+b = -5.

This is where I ran into difficulties - I think there is no way to solve for a and b using elimination, since both equations generated above are identical, and I tried "isolating" a as well as b from the equation a+b=-5, and plugging this back into one of the equations (ex. 4x-1=ax^2+(-5-a)+8 and substitue x=1), but this only generates tautologies such as 3=3.

I also tried plugging different values for a and b into a graphing program, to no avail (and ideally I'd still need to find an algebraic proof too.)

Is there something I'm missing here? Is there possibly no single solution to this problem? Or, another way to solve for a and b?
In situations such as this (fewer equations than unknowns) there are an infinity of solutions (or none). In this case let a be what it will and set b to -(a+5). Thus you have
f(x) =
ax^2 - (a+5) for x<-1
4x-1 for -1≤x<1
ax^2 - (a+5) x + 8 for x≥1

Of course you could have done it the other way, that is picked b.
 
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