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Mathematics 7 Online
OpenStudy (anonymous):

Function Notation problem Let f(x)=2x+1. Is f(3+1)=f(3)+f(1)? please answer it ???? thank you

OpenStudy (vishweshshrimali5):

No

OpenStudy (vishweshshrimali5):

See: for any function f(x), f(a+b) means that in the formula of f(x), put x = a+b.

OpenStudy (vishweshshrimali5):

For example; let, f(x) = 5x + 4 Then, f(2) = 5*2 + 4 = 14 f(5) = 5*5 + 4 = 29 f(2+5) = f(7) = 5*7 + 4 = 39 \(\ne\) f(5)+f(2)

OpenStudy (anonymous):

how to answer it

OpenStudy (vishweshshrimali5):

You can answer it in this way: First calculate f(3+2) i.e. f(5), then calculate f(3) and f(2), then show that f(5) \(\ne\) f(2) + f(3).

OpenStudy (vishweshshrimali5):

As I did in above example.

OpenStudy (anonymous):

my question is a example

OpenStudy (anonymous):

how to answer it this my question??

OpenStudy (anonymous):

Let f(x)=2x+1. Is f(3+1)=f(3)+f(1)?

OpenStudy (vishweshshrimali5):

Okay here is the detailed solution.

OpenStudy (vishweshshrimali5):

Since, f(x) = 2x + 1, thus, f(3) = 2*3 + 1 = 6 + 1 = 7 ---------(1) f(1) = 2*1 + 1 = 2 + 1 = 3 ---------(2) f(3+1) = f(4) = 2*4 + 1 = 8 + 1 = 9 ---------(3) Now, f(3) + f(1) = 7 + 3 = 10 ------------ (4) Clearly, \[\large{f(3+1) \ne f(3) + f(1)}\]

OpenStudy (vishweshshrimali5):

Do you get this ?

OpenStudy (anonymous):

yes thanks

OpenStudy (vishweshshrimali5):

No Problem. :)

OpenStudy (anonymous):

wait

OpenStudy (anonymous):

last

OpenStudy (anonymous):

let f(x)= \[\sqrt{x}\] Is f(3+1)=f(3)+f(1)?

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