Need Help, medal + fan:
Which are true of the function f(x) = 49 (1/7)x? Check all that apply.
A:The domain is the set of all real numbers.
B:The range is the set of all real numbers.
C:The domain is x > 0.
D:The range is y > 0.
E:As x increases by 1, each y-value is one-seventh of the previous y-value.
F:The initial value for the function is

- anonymous

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- UsukiDoll

\[\large f(x)=49(\frac{1}{7})^x \]
is this the function for the question?
or
\[\large f(x)=49(\frac{1}{7})x \]

- anonymous

first one

- UsukiDoll

ok... well our function doesn't have denominators so there are no restrictions.

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## More answers

- UsukiDoll

the domain is at the x-axis
the range is at the y-axis

- UsukiDoll

we need to graph this function t-t

- anonymous

Okay..

- butterflydreamer

note: you can also re-write your function as:
|dw:1436850224394:dw|

- UsukiDoll

we need x = 1,2,3... and so on..
for
\[\large f(x)=49(\frac{1}{7})^x \]

- UsukiDoll

we can use desmos. easier XD

- anonymous

Ohh okay you have to graph to see it..

- UsukiDoll

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- anonymous

I wasn't graphing it, so that is why it was really difficult to see

- UsukiDoll

yeah... sometimes questions are easier when you graph but since this function is too screwed up (I don't feel like doing it manually lol) we use desmos which graphs it for us

- anonymous

So from the graph, I already concluded what some of the answers could be, but still not sure if I am right, could you check to see? I will screenshot it

- UsukiDoll

sure

- anonymous

##### 1 Attachment

- UsukiDoll

The first 2 is indeed all real numbers
as I mentioned earlier we don't have a function that's like a fraction
ex. \[f(x) =\frac{1}{x-2} \] which means we can't have x be 2

- anonymous

Right, so A and B are correct..

- UsukiDoll

yes... hmm interesting on the last one.. I mean that would work if x = 1, but I could've sworn that initial value would mean that x = 0 .

- UsukiDoll

like.. if x = 1
\[\large f(1)=49(\frac{1}{7})^1\]
\[\large f(1)=49(\frac{1}{7})\]
\[\large f(1)=(\frac{49}{7})=7\]
If x = 0
\[\large f(0)=49(\frac{1}{7})^0\] (due to the zero exponent rule.. anything to the 0 results in 1 )
\[\large f(0)=49(1) = 49\]

- anonymous

Okay, we are getting somewhere haha. So we already know A and B are correct, anything else I am missing, or something I marked incorrectly?

- UsukiDoll

\[\large f(2)=49(\frac{1}{7})^2\]
\[\large f(2)=49(\frac{1}{49})\]
\[\large f(2)=1\]

- UsukiDoll

when my x is increasing I have
f(0) = 49
f(1) = 7
f(2) = 1

- anonymous

So Y is decreasing..

- UsukiDoll

I'm confused at that wording though... for the second to the last .. but yeah my y is going down.. fast.

- UsukiDoll

x y
0 49
1 7
2 1

- UsukiDoll

and it claims that when x is increasing the previous y is 1/7 th of the previous y value.

- UsukiDoll

?!

- anonymous

Right...? That is not true though. so we can eliminate that one

- UsukiDoll

I don't think so
it's like
49 ./ 7 =7
7/7 = 1
Then again
maybe if I calculate f(3)
\[\large f(3)=49(\frac{1}{7})^3 \]
\[\large f(3)=(\frac{49}{343}) \]
\[\large f(3)=(\frac{1}{7}) \]

- UsukiDoll

let's try f(4) maybe that question is talking about that?
\[\large f(4)=49(\frac{1}{7})^4\]

- anonymous

Hey I'm sorry I am running really short on time, So lets just sum it up. What do you think are the answers?

- UsukiDoll

\[\large f(4)=\frac{49}{2401}\]

- UsukiDoll

\[f(4) = \frac{1}{49}\]

- UsukiDoll

first two are fine
last one is when x = 0 , so it's not 7

- anonymous

Okay, and that is all?

- UsukiDoll

ugh trying to think to those 3 ...

- UsukiDoll

second to the last one is confusing... x.x

- anonymous

We'll just say no to that one...SO now we are left with C and D.

- UsukiDoll

if it's all reals then that would have the domain and range at >0

- UsukiDoll

I see it well on the range though. when y > 0

- UsukiDoll

on the graph

- UsukiDoll

but the domain is smoosh... like it's just on the line

- UsukiDoll

so first two are fine
domain is smoosh it's just 0 flat
range is greater than 0
no idea - split on that one
it's not 7 because x = 0 for initial conditions.

- UsukiDoll

I think after x = 3 the previous is 1/7 due to
1/7
(1/7)(1/7)
(1/7)(1/7)(1/7)

- anonymous

Okay thank you

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