Question

Find a pattern, I can't

Answer 1

The function f has the property f(x)=f(x+1) for all numbers x. If f(4)=17, what is f(8)?

Answer 2

So I guess if f(3), then it really equals 4, for example.

Answer 3

plug x=4 in the given equation !
and tell me what do you get from there ?

Answer 4

So it is f(4+1) = 5

Answer 5

nope,it would be like this
\[\large \bf f(4)=f(5)=17\]

Answer 6

right?

Answer 7

I'm trying to follow you. Thanks

Answer 8

@mayankdevnani

Answer 9

sorry! was helping another user

Answer 10

okay! similarly,find f(6)

Answer 11

Um. So would it be f(7)=18?

Answer 12

I'm not trying to be rude, or just ask for the answer, but I'm clueless. Could you just write it out? I'm going nowhere.

Answer 13

for finding f(6),just plug x=6 in given equation,we get
\[\large \bf f(6)=f(7)=17\]

Answer 14

similarly, find f(7) and finally f(8)
so you'll notice that we would get f(8)=17

Answer 15

That is weird. Ok. I do not see the use in that at all but ok.

Answer 16

so basically it says f(anything)=17

Answer 17

yep!

Answer 18

Hum. Thanks so much!

Answer 19

np !

Answer 20

@AmTran_Bus if you still need a little more convincing...
\[f(4)=17 \\ f(n)=f(n+1) \\ \]
this above stuff is the given
I'm going to write k equations next:
\[f(4)=f(5) \\ f(5)=f(6) \\ f(6)=f(7) \\ f(7)=f(8) \\ f(8)=f(9) \\ \cdots \\ f(k-1)=f(k) \\ f(k)=f(k+1)\]
now if a=b and c=d then a+c=b+d
so I'm going to put my k equations together...
\[f(4)+f(5) +f(6)+f(7) +f(8)+ \cdots +f(k-1)+f(k) \\ =f(5)+f(6)+f(7)+f(8)+f(9)+\cdots +f(k)+f(k+1)\]
now if I try to put everything on one side you should see a lot of cancellation

\[f(4)-f(k+1)=0 \\ \text{ so } f(k+1)=f(4) \\ \text{ but } f(4)=17 \\ \text{ so } f(k+1)=17 \\ \text{ so } f \text{ is a constant function for } \\ \text{ any integer } k \\ \text{ I could say } f(x)=17 \text{ for any } x \text{ that is an integer }\]

Answer 21

now in my opinion not all of this work is needed
i just wrote all of this if you by chance would find it helpful

Answer 22

another way to think about it
\[\frac{f(n+1)-f(n)}{n+1-n}=0 \text{ means the slope of } f \text{ is } zero \\ \\ \text{ which means } f \text{ is a constant function }\]

Answer 23

that equation I wrote is still your equation

Answer 24

\[\frac{f(n+1)-f(n)}{n+1-n}=0 \\ \frac{f(n+1)-f(n)}{1}=0 \\ f(n+1)-f(n)=0 \\ f(n+1)=f(n)\]

Answer 25

I think it would be helpful if you understood function notation better.

\(f(n) \) means the value of the function evaluated at \(x = n\).
In other words, the y-coordinate of the point where \(x = n.\)

You are told that \(f(n) = f(n + 1).\)
That means that the y-coordinate of the point where \(x = n\) is the same as the y-coordinate of the point where \(x = n + 1\).

Let's let x = 4.
Then f(4) = f(4 + 1) = f(5)
This means that the y-coordinates of the3 points that have x = 4 and x = 5 are the same.
If you let x = 5, then f(5) = f(5 + 1) = f(6)
Once again, the y-coordinates of the points that have x = 5 and x = 6 are the same.

You are told that f(4) = 17.
By letting x = 4, you see that f(5) = 17
By letting x = 5, you see that f(6) = 17
You keep on going until you do f(8), and you will see that y = f(8) = 17.

Answer 26

you are probably right @mathstudent55
I see there might be interpretation problems given amtran has written
"So it is f(4+1) = 5"

Answer 27

by the way instead of adding those k equations I wrote
you could use the property of transitivity
that is
if a=b and b=c then a=c

so for example
we had f(4)=f(5) and f(5)=f(6) so f(4)=f(6)

and by the substitution property of equality
if f(4)=17 then f(5)=17 and f(6)=17

and so on....

Answer 28

basically f(n)=f(n+1)
means term equals next term

Answer 29

I agree with @mathstudent55

But I think there's a shorter way to explain this

f(x) = f(x+1) essentially tells you one y-value is equal to the next y-value

The only way for that to be true... is if they're all equal to the same y-value.

Answer 30

U can also think of it like this-
Consider y=f(x)
Then the slope of this function between any two points f(x) and f(x+1) will be-
[f(x+1)-f(x)]/[(x+1)-x]

=0/1=0

Since the slope is 0 you can say that f(x) is not varying with x so the value of f(x) remains constant

Answer 31

^ that works too.