I was asked by a teacher to .. "Try to simply calculate 100(f[x+1]/f[x]-1) without plugging in any of x=0 or x=1 to see if it is a function of x."
What does he mean by, "To see if it is a function of x" ?
and "To simply calculate"

- anonymous

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- TheSmartOne

@dan815

- anonymous

and what does he mean by 'simply calculate' ?

- anonymous

was he talking about a particular expression for f(x) ?

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## More answers

- anonymous

oh yeah,., the f[x] = 5.1x^3
Does he just want me to fill in the equation ?

- anonymous

and simplify it?

- Australopithecus

There is a thing called a function test.
If you can draw a horizontal line anywhere through the graph of a function and if it passes through that function more than once it is not a function
For example:
This is a function a line only can pass through it once
|dw:1433516128053:dw|
This is not a function a horizontal line can pass through it more than once
|dw:1433516235183:dw|

- anonymous

so a vertical test .. that would prove / disprove if it was a function or not.. I suppose that's a good start.. does it prove if it is a function of x though?

- Australopithecus

100(f[x+1]/f[x]-1)
\[100*(\frac{f(x+1)}{f(x)} -1)\ or\ 100*(\frac{f(x+1)}{f(x)-1})\]

- anonymous

is that a mathematical way of proving that there is precisely only 1 solution for each value of x ?

- Australopithecus

yes

- anonymous

Im thinking maybe he means he wants me to expand and cancel out the equation to see if it just leaves x ?

- anonymous

or a more simple graph

- Australopithecus

Well you don't know the identity of f(x) or f(x+1) so you cant graph them. I have never dealt with a question such as this so maybe someone else could help you.

- anonymous

I dont think the vertical line test is the answer here..
Let me explain.. I had 3 problems to solve.. and they looked like this
"
Measure the average percent growth rate of
f[x] = 5.1 x^3
as x advances by 1 unit.
Does your result depend on where x starts and stops as it advances by one unit?
"
I used the same method to answer all 3.. I just plugged 0 and 1 in for x, and measured the change over 1 unit ..
The first two problems came back correct. The 3rd was a fail, and he asked me to 'simply calculate and prove that the function is dependent on x' etc.. I am on my 3rd attempt to get it right.. and I have no idea what the hell he is asking.. but I know it is not a vertical line test.

- Australopithecus

so f(x) is just 5.1x^3?

- anonymous

yes on that... I think maybe he wants me to prove algebraically that the percentage change of f(x+1)/f(x) will never be of the same proportion if x changes maybe? and therefore the output is dependent on x ? Im, not sure that follows actually.. I suppose a definition of dependency is in order.. I mean under what conditions is a function not dependent upon it's variable? I would guess, when it always returns the same value no matter what x=???, or if it can be shown that the output does not change for different values of x, under some condition maybe? I mean if a function will always return a different value if x changes, this must prove dependency right?
would you fall back then on the laws that govern limits, so you can throw away parts of the equation, like constants? Because constants will not affect the output in relation to x? And then show that the equation breaks down to some simpler equation?

- anonymous

okay, let me ask again.. what does it mean by, "to see if it is a function of x" ?

- Australopithecus

From my understanding a function would not be dependent on x when it is constant, so for instance.
f(x) = 3
the graph will be constant
|dw:1433551698364:dw|
5.1x^3 = f(x) definitely depends on x, not sure what your teacher is trying to demonstrate with:
100(f[x+1]/f[x]-1)
Just plug in and simplify I dont know maybe @ other people get their opinion on this.

- Australopithecus

All I know that the definition of a function is an equation that has for every input a unique output at least for non multi input functions

- Australopithecus

Or you can ask your teacher to clarify the purpose of this exercise

- kirbykirby

I just thought I'd add my 2 cents. While the vertical line test can tell you if you have a function or not, it is maybe just not enough of a "proof" to know whether or not it is actually a function.
My guess is as good as what was already mentioned, try to fill in the given expression using your \(f(x)=5.1x^3\) and verify whether or not the simplifcation leaves and x or not. As @ 86
Australopithecus mentioned, if you end up with a constant, it won't be a function of x.
If there is \(x\) remaining in the expression, then it *should* be okay to say that you have a function of x, but it should be checked whether only one value of y is assigned for any x.

- kirbykirby

Maybe something that might be enlightening if to consider what is not a function. Take the circle for example (say \(x^2+y^2=1\) ); it is not a function because you will have two possible y-values values for a single x-value and this occurs a least for one point. In particular, this will happen for all points where \(x \in (-1,1\), that is the open interval. At exactly \(x=-1\) and \(x=1\), there is only 1 y-value.
If you solve for \(y\), you get:
\(x^2 + y^2 = 1\\
y^2 = 1-x^2\\
y=\pm \sqrt{1-x^2}\)
so, as you can see, the fact that you have a \(\pm\) indicates that for "all" values of \(x\) (except at -1 and 1), you will have two possibly y-values for each x-value)

- kirbykirby

**\(x \in (-1, 1)\)

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