Limit Question

1. Oct 3, 2004

pulau_tiga

Find the constants a and b such that the function is continuous on the entire real line:
f(x) = {e^ax , x < or = to 1 ; 2x+b, x >1

I found the limits as x approaches 1 from the left and the right.
and set the two equations equalling each other
e^a = 2+b

However, I do not know what to do now. I have two unknowns.
If anyone could help me, or point me in the right direction. It would be greatly appreciated.
Thanks.

2. Oct 3, 2004

Motifs

No, there is no need to find the exact a and b. These two can be anything, any numbers to make the given function continuous, which i think is also what the problem's main question.
e^a-b=2 might be the answer ! Is that correct ?

3. Oct 3, 2004

e(ho0n3

You could let a = 0 so f(x) = 1 for x ≤ 1 and b = -1 so f(x) > 1 for x > 1. I'm a little concerned about the "continuous on the entire real line" bit since it implies (at least to me) that the range of f(x) is the reals which is impossible because you can never cover all of the negative reals.

4. Oct 3, 2004

Motifs

I might misunderstand what you meant but the given function is always beautifully POSITIVE REALS, there are actually NO negative reals at all. Any values of a and b given by someone can of course substitute immediately in the functions f(x) and it depends on the values of x's for people to choose the correct function to go on computing the POSITIVE REAL VALUEs of f(x)....

// Is that correct again ? :bigsmile:

5. Oct 4, 2004

maverick280857

The wording of the problem deserves some comments. The phrase, "the entire real line" indeed means that the function is to be made continuous over the domain (or if you like, interval) of values of x for which it is defined. It is however defined for all real x as you can see from the very definition of f(x). Hence, to make it continuous over the real line--which by the way represents the values of the independent variable by convention and not the those of f(x) which will be positive in this case--you need to equate the left and right hand limits.

Firstly if you can find a constraint relating the two 'constants' a and b (as you have rightly done), you have a perfectly valid solution.

Secondly, the function is defined differently for x<=1 and for x>1. So x = 1 is perhaps the only point where a discontinuity might arise (and it indeed will if e^a - b does not equal 2 for a particular pair of values of (a,b) ).

Finally, if the question specifically requires you to state a particular set (a,b) then you can set a = 1 and b = e - 2. All that simple really :-). Note that if you write e^a = b + 2, you are saying (though not explicitly) that the right hand side, viz. b + 2 is positive. This means that b > -2. So long as b satisfies this criterion, there are an infinite number of values of a (do you see why?). Now if b is nonnegative, then the constraint amounts to saying that the right hand side is at least 2.

Alternatively if you fix a, you have fixed e^a and so b = e^a - 2 is like b = y - 2 (y = e^a) so for every (positive value of y) there is a unique value of b. However, there are an infinite number of positive real values that y can take, so there are corrrepondingly infinitely number of real values (not necessarily all positive though) that b can take. It really depends on what the problem wants you to do now.

Hope that helps...

Cheers
Vivek