Expert: Ahmed Salami - 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
Hi Maureen,
You keep talking about 'functions' whereas there is in fact just ONE function, it just has two parts.
f(x) = {e^ax for x ≤ 1
    &  b + ln x  for x > 1}
By definition, a function is differentiable only at points where it is continuous. That said, but continuity alone is not a sufficient condition for differentiability.
Mathematically, a function f is differentiable at a point p if
lim   [f(p+h) - f(p)]/h
h→0
exists.
Technically speaking, a function would be differentiable at a point if the slopes, i.e looking at it from both right and left, both approach this same value.

But first we need to find condition for continuity at x = 1.
A function f(x) is continuous at x = a if the limit exists there and is equal to the function value at the point, i.e
lim x→a f(x) = f(a)
Simply put, at x = 1 the two parts of f(x) must have the same defined value i.e
e^ax = b + ln x    (at x = 1)
e^a(1) = b + ln (1)
e^a = b + 0
b = e^a
and that is the condition for continuity at x = 1
For differentiability, the derivatives need to be the same.
d(e^ax)/dx = ae^ax
d(b + ln x)/dx = 1/x
They need to be equal at x = 1, and so
ae^a(1) = 1/(1)
ae^a = 1
e^a = 1/a
Now, this kind of equation is easily solved by the graphical method. You can draw the graphs of e^x and 1/x with the x-value of the point of intersection being the value of a. And then,
b = 1/a
The graphical method or any one of the techniques of numerical methods solves the equation e^a = 1/a to give a = 0.5671
b = 1/0.5671
 = 1.7632


Hope it helps.
I've only tried to redo it from the beginning in the hope that it might make for better understanding.
You can always get back to me.

Regards