Expert: Abe Mantell - 12/28/2009

Question
The problem gives two functions with variables.  We are supposed to compare the equations and then compare the derivatives in order to get two equations with the two variables in them.  It then becomes an algebra problem (or trig) for identifying the value for each of variable that would make the two functions continuous and differentiable.

Two formulas:

f(x) = e^ax     if x is less than or equal to 1
f(x) = b + ln x if x is greater than 1

Can we find values of a and b that make the functions differentiable at x = 1?

I think we are getting especially hung up on the part that the answer key says "solve by the grapher."  The equation we came up with states that ae^a = 1.  I cannot get the number to equal 0.5671... (the book value) and I do not know how to use my grapher TI-84 plus to find this value.  The b value they get is 1.7632.

Any help you can give me would be great!

Thank you,

Maureen Johnson

Answer
Hello Maureen,

Let me just go through the entire problem.
So, we first need continuity at x=1, thus: e^a = b
Then we need differentiability at x=1, thus ae^a = 1
Good?

Since ae^a=1 cannot be solved analytically, graphically is a good way.
One way is to the y=xe^x and y=1, and see where they intersect...
or since ae^a=1 can be written as ae^a-1=0, we can just graph y=xe^x-1
and see where it has a root.  Either way you should get x is about 0.5671432904

Thus, a is about 0.5671432904, and b=e^a or about 1.763222834

OK?

I hope this helps.  Just curious, what class is this for?  High School calculus
or college?

Have a happy new year!

Abe

BTW: I will be soon be going on vacation from allexperts.com...if you have any other
    questions, you can contact me at mantell@ncc.edu