Question

The function f(x) = 20(3)^x represents the growth of a fish population every year in a local lake. Jesse wants to manipulate the formula to an equivalent form that calculates every half-year, not every year. Which function is correct for Jesse's purposes?

Answer 1

@jim_thompson5910

Answer 2

\(f(x) = 20(3^x)\)

Is this the correct function?

Answer 3

no. It is f(x)=20(3)^x. Like a log function :)

Answer 4

`Which function is correct for Jesse's purposes?`
sounds like you're given a list of choices

Answer 5

yes :)

Answer 6

This function looks like it triples the number of fish each year.

Answer 7

f(x)=20(3^2)^x/2
f(x)=20(3^1/2)^2x
f(x)=20/2(3)^x
(f(x)=40(2)^x
these are the answer choices

Answer 8

Let's look at the function from year 0 to year 3.

Year 0: \(y = 20 \cdot 3^0 = 20 \cdot 1 = 20\)
In year 0 there are 20 fish.

Year 1: \(y = 20 \cdot 3^1 = 20 \cdot 3 = 60\)
In year 1 there are 60 fish.

Year 2: \(y = 20 \cdot 3^2 = 20 \cdot 9 = 180\)
In year 2 there are 180 fish.

Year 3: \(y = 20 \cdot 3^3 = 20 \cdot 27 = 540\)
In year 3 there are 540 fish.

As you can see, with each passing year, the population of fish triples.

Answer 9

Yup! I see that! Continue please

Answer 10

Now look at the given formulas.
Since you are now calculating the growth half a year, you again start with a population of 20.
After two half-years (which is 1 year), you must end up with 60.

Try each choice with x = 2 (meaning 2 half-years = 1 year) to see which formula gives you 60.

Answer 11

So would it be B?

Answer 12

From what I wrote above, you know that
f(0) = 20
f(1) = 60
f(2) = 180
f(3) = 540
Try each choice to see which one will give you those values, respectively, at
f(0)
f(2)
f(4)
f(6)

Answer 13

Choice B.
\(f(x)=20(3^{\frac{1}{2}})^{2x} = 20(3^x)\)

\(f(0) = 20 \cdot 3^0 = 20\)

\(f(2) = 20 \cdot 3^2 = 180\)

It turns out that choice B s the same as the given function, so that is not it.

Answer 14

Choice A.

\(f(x)=20(3^2)^{\frac{x}{2}} = 20 \cdot 3^x\)

Once again, just like choice B, choice A is also the same as the original function.

Answer 15

Then that leaves choice C and D.

Answer 16

So, C looks to be correct

Answer 17

Choice C.

\(f(x)=\dfrac{20}{2}(3)^x = 10(3^x) \)

f(0) = 10

This cannot be correct because it gives us an initial value different from f(0) = 20

Answer 18

Choice D.

\(f(x)=40(2)^x\)

\(f(0) = 20\)

Once again, choice D also cannot be it because the initial value is wrong.

Answer 19

so all of them are wrong??

Answer 20

Since none of the choices you listed worked, I have a question for you.
Notice that the functions in choices A and B were the same as the original expression.
Are you sure you copied them correctly?