Solve Continuity Problem: Find a and b | Homework Help

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Homework Help Overview

The problem involves determining the values of a and b to ensure the continuity of a piecewise function defined differently over various intervals. The function is given as ((x)^4-4)/(x-2) for x<2, a(x)^2-bx+3 for 2

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conditions for continuity at specific points (x=2 and x=3) and the need for limits to match function values at those points. Some suggest calculating left-sided and right-sided limits, while others question the definitions of the function at those points.

Discussion Status

There is ongoing exploration of potential values for a and b, with some participants proposing specific values and others questioning their validity based on the continuity definition. The discussion reflects uncertainty about the function's behavior at the discontinuities.

Contextual Notes

Participants note that the function is undefined at x=2 and x=3, raising concerns about the continuity conditions and the possibility of removable discontinuities. There is mention of the need for limits to be equal and the implications of different combinations of a and b on continuity.

realism877
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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)...
 
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]
 
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 ?
 
  • #10
I did that. I just want to know if I'm right.

A=1/2 b=1/2
 
  • #11
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.
 

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