Solve Continuity Problem: Find a and b | Homework Help

[realism877]

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

 

[micromass]

Hi realism877,

A function f is continuous in a if and only if \lim_{x\rightarrow a}{f(x)}=f(a). 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)...

 

[SammyS]

First of all, f(x) is undefined at x=2 and x=3 .

Once the above is fixed, you will need \lim_{\,x\to2^-}f(x)=\lim_{\,x\to2^+}f(x)=f(2) and \lim_{\,x\to3^-}f(x)=\lim_{\,x\to3^+}f(x)=f(3) \,.

 

[cloud360]

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

 

[realism877]

Is a=1/2 and b=1/2?

 

[HallsofIvy]

First of all, f(x) is undefined at x=2 and x=3 .

Once the above is fixed, you will need \lim_{\,x\to2^-}f(x)=\lim_{\,x\to2^+}f(x)=f(2) and \lim_{\,x\to3^-}f(x)=\lim_{\,x\to3^+}f(x)=f(3) \,.

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.

 

[HallsofIvy]

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.

 

[realism877]

I know what coninuity means, but we are asked to to solve for a and b.

I did that and I got those values.

 

[QuarkCharmer]

Can't you somehow turn this into a system of equations with 2 unknowns and then use substitution/elimination to solve a/b ?

 

[realism877]

I did that. I just want to know if I'm right.

A=1/2 b=1/2

 

[SammyS]

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.