Solve Continuity Problem: Find a and b | Homework Help

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Homework Statement


I have to find out what a and b is to make it continuous everywhere

((x)^4-4)/(x-2) if x<2a(x)^2-bx+3 if 2<x<3

2x-a+b if x greater than or equal to 3

Homework Equations


I don't know what I'm doing to solve this problem.

The Attempt at a Solution

 
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Hi realism877, :smile:

A function f is continuous in a if and only if [tex]\lim_{x\rightarrow a}{f(x)}=f(a)[/tex]. In particular, the right-sided limit must equal the left-sided limit.

So, in your example, you must calculate the left-sided limits and the right-sided limits in 2 and 3 and make sure they are equal to f(2) and f(3)...
 
Usually for these questions, you need to make sure that the limits on both sides are the same. e.g 0.00001 and -0.0001 don't jump i.e limit does not change
 
Is a=1/2 and b=1/2?
 
SammyS said:
First of all, f(x) is undefined at x=2 and x=3 .

Once the above is fixed, you will need [tex]\lim_{\,x\to2^-}f(x)=\lim_{\,x\to2^+}f(x)=f(2)[/tex] and [tex]\lim_{\,x\to3^-}f(x)=\lim_{\,x\to3^+}f(x)=f(3) \,.[/tex]
No, f(3) is defined as 6- a+ b. But you are right that the way the problem is given f(2) is not defined and no values of a and b will make the function continuous there.
 
realism877 said:
Is a=1/2 and b=1/2?
Why are you asking? Do you know what "continuity" means? If a= 1/2 and b= 1/2 will this function satisfy the definition of "continuity" at x= 2 and x= 3.
 
I know what coninuity means, but we are asked to to solve for a and b.

I did that and I got those values.
 
Can't you somehow turn this into a system of equations with 2 unknowns and then use substitution/elimination to solve a/b ?
 
I did that. I just want to know if I'm right.

A=1/2 b=1/2
 
realism877 said:
Is a=1/2 and b=1/2?
Well, that combination does make the function continuous at x = 3 (if you define f(3)=6), but so do many other combinations of a & b. For instance, a = 3/2 & b= 3 makes f continuous at x = 3, if you define f(3) = 15/2 .

The discontinuity at x = 2 is NOT removable - unless there is a typo in the definition of f(x) for x < 2. There is no factor in the numerator to cancel the factor of (x - 2) in the denominator.