Question

For which of the following functions does f(a)+f(b) = f(a+b)?
A: y=2x+3
B: y=2x
C: y=x^2
D: y=sqrtx
E: y=|x|

Answer 1

@myininaya

Answer 2

Did you do f(a)+f(b) and f(a+b) for each of the following

Answer 3

any linear function

Answer 4

Yes, and I'm not sure which one it would be..

Answer 5

is it y=2x? :)

Answer 6

y=x^2
is parabola
y=sqrtx
is curve
y=|x|
is like inverted triangle:

so what's left are lines

Answer 7

that one is linear
and you see if f(a)+f(b)=f(a+b)

Answer 8

f(x)=2x
f(a)=2a
f(b)=2b
and
f(a+b)=?
Then you will see that f(a+b)=f(a)+f(b)
after a little distribution

Answer 9

But

Answer 10

There is also another one

Answer 11

there is?

Answer 12

no only one sorry

Answer 13

Ok so it is y=2x then?:)

Answer 14

yep

Answer 15

it has to be linear but you have to watch out to see if has +constant other than 0

Answer 16

can you help with this? The domain of f(x)=sqrt(x-2)/x^2-x is?

Answer 17

\[f(x)=\frac{\sqrt{x-2}}{x^2-x} ?\]

Answer 18

here are the options.. all reals except 0 and 1
x</= 2, x can't be 0 or 1

Answer 19

yes

Answer 20

I got all reals, but it can't be 0 or 1.. but is is also x< or equal to 2?

Answer 21

Ok the thing under the radical needs to be 0 or positive so that means we need
x-2 _?__ 0
What goes where that question mark is...
Less than?
Greater than?
Greater than or equal?
Less than or equal?
-------
The you also don't want the bottom to be zero

Answer 22

les than or equal? I'm not sure what you're asking

Answer 23

its less than or equal as an option

Answer 24

No we want x-2 to be zero or positive
so that means we want x-2> or = 0

Answer 25

\[x-2 \ge 0\]
Be we also want:
\[ x^2-x \neq 0\]

Answer 26

So would the answer be x</= 2, then can't be 0 and 1

Answer 27

I agree with myininaya

Answer 28

No no you want x-2 >=0
not <=

Answer 29

Oh ok so the answer would be x>= 0, 2?? thats it? i thought it can't be 0 and 1

Answer 30

You need to solve that inequality I wrote up there

Answer 31

yeah, its x is greater than or equal to 2.

Answer 32

right! :)

Answer 33

but just that then? or 0 and 2 can't be

Answer 34

0 and 2?

Answer 35

so thats just the answer? because after i factored, i found that 0 and 1 can't be it

Answer 36

x(x-1)=0, so it can't be 0 and 1..but its just x>=2?

Answer 37

you mean the bottom is 0 when x=0 or x=1 right
but since x is greater than or equal 2 then we don't have to worry about x=0 or x=1 because x is greater than these numbers

Answer 38

Ok so you're sure the answer will just be x>/= 2?

Answer 39

Am I sure?
Am sure if you are sure?
Do you need further explanation?
Like I mean what is bugging you about this answer?

Answer 40

I think thats the answer, but is it is what I'm asking? lol

Answer 41

Yes if the function is:
\[f(x)=\frac{\sqrt{x-2}}{x^2-x}\]
Then the domain is \[x \ge 2 \]