Hello, waunar!
An investment of P dollars increased to A dollars in t years.
If interest was compounded contiouously, find the interest rate.
A = 13,464, P = 1000, t = 20, r = unknown
I can solve the problem if the they ask me for A p or t but i don't know how to solve for r.
Hmmm, you say you know hot to solve for \(\displaystyle t\), right? . . . I don't see any problem here.
We have: \(\displaystyle \:A\:=\
e^{rt}\;\;\Rightarrow\;\;13,464\:=\:1000e^{20r}\)
Use the same method you used to solve for \(\displaystyle t\) . . .
Divide by 1000: \(\displaystyle \;e^{20r}\:=\:13.464\)
Take logs: \(\displaystyle \;\ln(e^{20r})\:=\:\ln(13.464)\;\;\Rightarrow\;\;20r\cdot\ln(e)\:=\:\ln(13.464)\;\;\Rightarrow\;\;20r\:=\:\ln(13.464)\)
Hence: \(\displaystyle \:r\:=\:\frac{\ln(13.464)}{20}\:=\:0.130000973\)
\(\displaystyle \;\;\)Therefore: \(\displaystyle \,r\:=\:13\%\)
Graph \(\displaystyle f\) on the interval (0, 200).
Find an approximate equation for the horizontal asymptote.
\(\displaystyle \;\;\;f(x)\;=\;\left(1\,+\,\frac{1}{x}\right)^x\)
Are we expected to look at the graph and "eyeball" the asymptote?
\(\displaystyle \;\;\)Kind of primitive, ain't it?
A horizontal asymptote occurs if \(\displaystyle \,\lim_{x\to\infty}f(x)\) has a finite limit.
It happens that: \(\displaystyle \,\lim_{x\to\infty}\left(1\,+\,\frac{1}{x}\right)^x\,\) is the
definition of \(\displaystyle e\).
Therefore, the horizontal asymptote is: \(\displaystyle \,y\:=\:e\)
