Question

Find (f + g)(x).

Answer 1

\[ g= \sqrt{6x-9} \]

Answer 2

f=\[\sqrt{6x+9}\]

Answer 3

\[(f+g)(x)=f(x)+g(x)\]
Use this formula, everything else is given

Answer 4

what subject is this for?

Answer 5

precal

i think i got it is the answer 6x

Answer 6

I believe so.

Answer 7

can you help me with one more

Answer 8

sure

Answer 9

Determine algebraically whether the function is even, odd, or neither even nor odd.
f(x)= x+ 4/x

Answer 10

what are the answer options just want to check

Answer 11

even odd or neither

Answer 12

f(x)= x+4/x in my opinion would be odd mainly because its the opposite of the 4/x

Answer 13

excuse me if im wrong i havent been studying this

Answer 14

(f+g)(x)=f(x)+g(x) don't know how you got 6x if really is that
\[f(x)=\sqrt{6x+9} \text{ and } g(x)=\sqrt{6x-9}\]

Answer 15

just replace f(x) with sqrt(6x+9)
and replace g(x) with sqrt(6x-9)

Answer 16

also to determine if a function is odd or even (or neither odd or even) the first step is to plug in -x

Answer 17

if you receive f(-x)=f(x), then f is even
if you receive f(-x)=-f(x), then f is odd

Answer 18

so it would be odd am i correct

Answer 19

\[f(x)=x+\frac{4}{x} \\ \text{ plug \in } -x \\ f(-x)=-x+\frac{4}{-x} \\ f(-x)=-(x+\frac{4}{x}) \\ f(-x)=-f(x) \\ \text{ yep } f \text{ is odd } \\ \text{ unless you meant } f(x)=\frac{x+4}{x} \\ \text{ then the story is a bit different }\]

Answer 20

you would still plug in -x of course

Answer 21

yes