Question

A certain bacteria population is known to triples every 150 minutes. Suppose that there are initially 110 bacteria. What is the size of the population after t hours?

Answer 1

You know it is an exponential functions when they say 'doubles', 'triples' for some constant amount of time.

Answer 2

Exponential functions come in the form: \[\Large ce^{kt}\]You have to use the information they give you to solve for \(c\) and \(k\).

Answer 3

may i have more details.... i thought the answer is fx=110*3^(t/2/5)..... am i right?

Answer 4

Usually something like 'initially' or 'at the start' is a good indicator for solving for \(c\). Since at that time \(t=0\).

Answer 5

t/2/5? Is that \(\frac{t}{2/5}\) or \(\frac{t/2}{5}\)?

Answer 6

the first one,

Answer 7

It looks good. Why don't you just check?

Answer 8

When \(t=0\) does it equal \(110\)?

Answer 9

Does \(\frac{f(t+2.5)}{ f(t)} = 3 \)?

Answer 10

These are ways you would check!

Answer 11

Does that make sense?

Answer 12

ok, i am double right now, thank you

Answer 13

What language do you normally speak?

Answer 14

yes, When t=0 does, the bacterial should
equal 110

Answer 15

chinese

Answer 16

why you are asking?

Answer 17

Maybe someone who speaks Chinese would see and help too. Who knows.

Answer 18

Now did you try to divide \(\frac{ f(t+2.5) }{ f(t) } \)?

Answer 19

you are right, the problem is not language, it's i just totally did not remember exponential question anymore

Answer 20

well,f(t+2.5)f(t) equals 3.
so, it's triple from 110 to 330, right?

Answer 21

Yeah

Answer 22

Let me see...

Answer 23

may i ask another question about cubic function?

Answer 24

\[\huge\frac{110\cdot 3^{\frac{t+2.5}{2.5}}}{110\cdot 3^{\frac{t}{2.5}}}\]\[\huge=\frac{110\cdot 3^{\frac{t}{2.5}+1}}{110\cdot 3^{\frac{t}{2.5}}}\]\[\huge=3\cdot \frac{110\cdot 3^{\frac{t}{2.5}}}{110\cdot 3^{\frac{t}{2.5}}}\]\[\huge=3\]So it's correct.

Answer 25

What about cubic function?

Answer 26

Find the equation for the cubic function, f(x), with roots at 5, 6 and -5, and has a y-intercept at (0, 13).

Answer 27

So, remember that \[(x-r_1)(x-r_2)(x-r_3)=0\]Where \(r\) are roots.

Answer 28

But even after that, we have to compress/expand it so that the y-intercept is correct.

Answer 29

so, what i got f(x)=x^3-6x^2-25x+13, I try soooo long, still got the same answer, and after I double check, the answer is wrong

Answer 30

You need to compress it

Answer 31

yesss. you are right about the root formular

Answer 32

Now, what is the current y intercept? What happens when \(x=0\)?

Answer 33

when x=0, y=13?

Answer 34

\[c(x-5)(x-6)(x-(-5))=0\]Okay so \(c\) is some constant... ok?

Answer 35

We need to solve for \(c\). It will compress our function.
To solve for \(c\), plug in the point \((0,13)\)
Like this:
\[c((0)-5)((0)-6)((0)-(-5))=13\]

Answer 36

Now can you solve for \(c\)?

Answer 37

ok...is c= 13/150 or c=13-150

Answer 38

c=13/150, right?

Answer 39

Yeah

Answer 40

So our function is just:
\[y=\frac{13}{150}(x-5)(x-6)(x+5)\]But you need to expand the polynomials.

Answer 41

yes, sounds great! now, i got it,

Answer 42

Do you understand how I did that?

Answer 43

I don't remember much anything about cubic or exponential functions. I did it by using rules of algebra.

Answer 44

yes, i indeed understood, thanks