1. ## exponential

if f(x) behaves exponentially and f(-1) = 8 and f(1) = 4

1) Find a formula for f(x) using the natural base e:
2) Find a formula for f(x) using another base suited to the data:
3) What is differenital equation satsfied by f(x)?
4) Where does f(x) = 1?

2. Originally Posted by viet
if f(x) behaves exponentially and f(-1) = 8 and f(1) = 4

1) Find a formula for f(x) using the natural base e:
2) Find a formula for f(x) using another base suited to the data:
3) What is differenital equation satsfied by f(x)?
4) Where does f(x) = 1?
$f(x) = Ae^{kt}$

Thus,

$4=f(1) = Ae^k$ (1)
$8=f(-1) = Ae^{-k}$ (2)

Divide (2) by (1):
$\frac{Ae^{-k}}{Ae^k} = \frac{8}{4} =2$
Thus,
$e^{-2k} = 2$
Thus,
$-2k = \ln 2$
Thus,
$k = -\frac{1}{2} \ln 2$

Now you can solve for A.

3. Hello, viet!

Here's part (2) . . .

If $f(x)$ behaves exponentially and $f(-1) = 8$ and $f(1) = 4$

2) Find a formula for $f(x)$ using another base suited to the data
I would use base 2 . . .

We have: . $f(x) \:=\:A\!\cdot\!2^{kt}$

$\begin{array}{ccc}\text{Since }f(\text{-}1) = 8: & A\!\cdot\!2^{\text{-}k} \:=\:8 & [1] \\
\text{Since }f(1) = 4: & A\!\cdot\!2^k \:=\:4 & [2]\end{array}$

Divide [2] by [1]: . $\frac{A\!\cdot2^k}{A\!\cdot\!2^{\text{-}k}} \:=\:\frac{4}{8}\quad\Rightarrow\quad 2^{2k} \:=\:\frac{1}{2} \:=\:2^{\text{-}1}\quad\Rightarrow\quad 2k = \text{-}1\quad\Rightarrow\quad k = \text{-}\frac{1}{2}$

The function (so far) is: . $f(x) \:=\:A\!\cdot\!2^{(\text{-}\frac{1}{2}t)}$

Since $f(\text{-}1) = 8:\;A\!\cdot\!2^{(\text{-}\frac{1}{2})(\text{-}1)} \:=\:8\quad\Rightarrow\quad A\!\cdot\!2^{\frac{1}{2}} \:=\:8\quad\Rightarrow\quad A \:=\:4\sqrt{2}$

The function is: . $f(x) \;=\;4\sqrt{2}\cdot2^{(\text{-}\frac{1}{2}t)}\;=\;2^2\cdot2^{\frac{1}{2}}\cdot2^ {(\text{-}\frac{1}{2}t)} \;=\;2^{(\frac{5}{2}-\frac{1}{2}t)}$

Therefore: . $\boxed{f(x) \;=\;2^{\frac{1}{2}(5-t)}}$