Unifying a Piecewise Function: Finding Values for Continuity

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SYoungblood
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Homework Statement


Hello, thank you in advance for all help. This is a limit problem that is giving me a particularly hard time.

Homework Equations


For what values of a and b is f(x) continuous at every x? In other words, how to unify the three parts of a piecewise function so that there are no holes or jumps in the function?

f(x) = { (i) -2 for values less than or equal to -1; (ii) ax - b for values of -1 < x < 1; and (iii) 3 for values greater than or equal to 1?

The Attempt at a Solution


[/B]
The lefthand limit for (i) as x approaches -1 and the righthand limit for (iii) as x approaches 1 are fairly straightforward. (ii) is a polynomial, so the limit is continuous over that interval. In theory, and most certainly in practice, there are values for the limits as x approaches -1 and 1 that make all of f(x) continuous. However, I cannot seem to reconcile how to substitute values for x to find a and b. If I set ax - b = -1, or 1, the way I see it, I have two variables and cannot account for both.

Thank you again,

SY
 
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SYoungblood said:
If I set ax - b = -1
This has to be true for a specific value of x only.
+1 is the result for a different specific value of x.

That gives you two equations with two unknowns.
 
SYoungblood said:

Homework Statement


Hello, thank you in advance for all help. This is a limit problem that is giving me a particularly hard time.

Homework Equations


For what values of a and b is f(x) continuous at every x? In other words, how to unify the three parts of a piecewise function so that there are no holes or jumps in the function?

f(x) = { (i) -2 for values less than or equal to -1; (ii) ax - b for values of -1 < x < 1; and (iii) 3 for values greater than or equal to 1?

The Attempt at a Solution


[/B]
The lefthand limit for (i) as x approaches -1 and the righthand limit for (iii) as x approaches 1 are fairly straightforward. (ii) is a polynomial, so the limit is continuous over that interval. In theory, and most certainly in practice, there are values for the limits as x approaches -1 and 1 that make all of f(x) continuous. However, I cannot seem to reconcile how to substitute values for x to find a and b. If I set ax - b = -1, or 1, the way I see it, I have two variables and cannot account for both.

Thank you again,

SY

Why would you set ax-b=(-1)? You want to set ax-b=(-2) when x=(-1) and ax-b=3 when x=1. Solving those two equation for a and b shouldn't be much of a problem.
 
If a function, f, is continuous for all x< a, then its limit as x approaches a from below is f(a). Similarly, if a function, g, is continuous for all x> a, then its limit as x approaches a from above is g(a). If a function is defined to be f(x) for x< a and g(x) for x>a, then it is continuous at x= a if and only if f(a)= g(a).
 
Thank you for the help Dick, I got it right! (After much weeping and gnashing of teeth.)

SY