Let f(x) = 3x - 6 and g(x) = x - 2. Find and its G/H domain
A. 3; all real numbers except x = 2 B. 3; all real numbers C. -3; all real numbers except x = 3 D. 1; all real numbers
what is the first thing we have to find ?
oppps find F/G and its domain
\[ \frac{3x - 6}{x-2} \] can you factor 3 from the expression in the numerator?
I have noo idea.. sorry I'm lost with all this stuff.. :/
The idea is like this: say you have two things a + b grouped together (i.e. added) and we have 3 of these groups: 3(a+b) 3(a+b) means 3 times (a+b) if you have 3 of these "packages" it should be more or less obvious we have 3 of each thing inside the package: 3(a+b) = 3a + 3b factoring "undoes" the 3a + 3b if you can evenly divide both terms by 3 you can rewrite 3a+3b as 3(a+b)
in your case, you have 3x -6 notice both terms can be divided evenly by 3
Right
can you try to factor 3 out of 3x - 6 ?
I'm not really sure how to write it because theres only an X without any numbers next to it.. soo idk how to factor them together..
3x/3 is x -6 /3 is -2 you get x-2 for the part that goes in the parens
3(x-2) ?
yes. you now have \[ \frac{3(x-2)}{(x-2)} \]
okay
I assume you know anything divided by itself is 1 (except 0/0 which we are not allowed to do)
right.. so the answer would be 3
(x-2) divided by itself is 1 the only exception is it is not allowed to be 0 in other words, we do NOT want x-2= 0 i.e. x=2 is not allowed
for all values of x except x=2, we get 3 (you can test a few values to see it really works) but for x=2 we run into a problem of dividing by 0. so we make a note: no divide by 0 or in other words, x ≠ 2
btw, the "domain" are all the x values we are allowed to use. we can use any number except x=2
Okay so were left with 1?
WAIT! I get it... itd be A? 3; all real numbers except x = 2
yes
lol thank you =]
yw
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