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

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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

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\[\large f(x)=49(\frac{1}{7})^x \] is this the function for the question? or \[\large f(x)=49(\frac{1}{7})x \]
first one
ok... well our function doesn't have denominators so there are no restrictions.

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the domain is at the x-axis the range is at the y-axis
we need to graph this function t-t
Okay..
note: you can also re-write your function as: |dw:1436850224394:dw|
we need x = 1,2,3... and so on.. for \[\large f(x)=49(\frac{1}{7})^x \]
we can use desmos. easier XD
Ohh okay you have to graph to see it..
I wasn't graphing it, so that is why it was really difficult to see
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
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
sure
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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
Right, so A and B are correct..
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 .
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 ) \[\large f(0)=49(1) = 49\]
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?
\[\large f(2)=49(\frac{1}{7})^2\] \[\large f(2)=49(\frac{1}{49})\] \[\large f(2)=1\]
when my x is increasing I have f(0) = 49 f(1) = 7 f(2) = 1
So Y is decreasing..
I'm confused at that wording though... for the second to the last .. but yeah my y is going down.. fast.
x y 0 49 1 7 2 1
and it claims that when x is increasing the previous y is 1/7 th of the previous y value.
?!
Right...? That is not true though. so we can eliminate that one
I don't think so it's like 49 ./ 7 =7 7/7 = 1 Then again maybe if I calculate f(3) \[\large f(3)=49(\frac{1}{7})^3 \] \[\large f(3)=(\frac{49}{343}) \] \[\large f(3)=(\frac{1}{7}) \]
let's try f(4) maybe that question is talking about that? \[\large f(4)=49(\frac{1}{7})^4\]
Hey I'm sorry I am running really short on time, So lets just sum it up. What do you think are the answers?
\[\large f(4)=\frac{49}{2401}\]
\[f(4) = \frac{1}{49}\]
first two are fine last one is when x = 0 , so it's not 7
Okay, and that is all?
ugh trying to think to those 3 ...
second to the last one is confusing... x.x
We'll just say no to that one...SO now we are left with C and D.
if it's all reals then that would have the domain and range at >0
I see it well on the range though. when y > 0
on the graph
but the domain is smoosh... like it's just on the line
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.
I think after x = 3 the previous is 1/7 due to 1/7 (1/7)(1/7) (1/7)(1/7)(1/7)
Okay thank you

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