# Calculus: Exponential Functions

• Sep 8th 2007, 04:34 PM
Deltaboy
Calculus: Exponential Functions
can someone help me with this problem..

http://i10.photobucket.com/albums/a1...Untitled-1.jpg

well, I thought that f(x)= 2*e^(-x) was pretty close except when x at 2, y is not (2/9) but instead it's .5

any one can help?
thanks..
• Sep 8th 2007, 04:43 PM
topsquark
Quote:

Originally Posted by Deltaboy
can someone help me with this problem..

http://i10.photobucket.com/albums/a1...Untitled-1.jpg

well, I thought that f(x)= 2*e^(-x) was pretty close except when x at 2, y is not (2/9) but instead it's .5

any one can help?
thanks..

Well it goes through (0, 2), so
$2 = Ca^0 \implies C = 2$

So the curve is $f(x) = 2a^x$.

Again, the curve goes through $\left ( 2, \frac{2}{9} \right )$ so,
$\frac{2}{9} = 2a^2$

$\frac{1}{9} = a^2$

Thus
$a = \pm \frac{1}{3}$

Now, we may exclude the negative solution because the curve for this is, to put it mildly, psychotic. Thus
$f(x) = 2 \left ( \frac{1}{3} \right ) ^x$

-Dan