Question

Let f(x) = 2-x and g(x) = 1/x. Perform each function operation and then find the domain of the result. (f-g)(x)
Show steps please, very confused

Answer 1

(f-g)(x) = f(x) - g(x) = (2-x) - (1/x) = x(2-x)/x - 1/x = (2x - x^2)/x - 1/x = (2x - x^2 - 1)/x

Answer 2

do you do the reciprocal of x/1?

Answer 3

(-x^2 + 2x - 1)/x = -(x^2 - 2x + 1)/x = -(x-1)(x-1)/x

Therefore, (f-g)(x) = -((x-1)(x-1))/x

Answer 4

Do you think you can do it in the equation thing, i get confused looking at it like that

Answer 5

\[(f-g)(x) = f(x) - g(x)\]\[f(x) - g(x) = (2-x) - (\frac{1}{x})\]

Answer 6

\[f(x) - g(x) = \frac{x(2-x)}{x} - \frac{1}{x}\]\[f(x) - g(x) = \frac{2x - x^2}{x} - \frac{1}{x}\]\[f(x) - g(x) = \frac{2x - x^2 - 1}{x}\]

Answer 7

why do you divide 2-x by x and multiply by x??

Answer 8

\[f(x) - g(x) = \frac{-x^2 + 2x - 1}{x} = -\frac{(x-1)(x-1)}{x}\]

Answer 9

We do that so that we get a common denominator of x to combine everything into one fraction. Remember, x/x = 1

Answer 10

the the domain would be all real numbers? is there any exceptions?

Answer 11

Maybe I should have wrote it like this so that it is easier for you to see:
\[\frac{x}{x} \times \frac{2-x}{1} = \frac{x(2-x)}{x}\]

Answer 12

\[x \ne 0\]

Answer 13

That literally means "x cannot equal zero"

Answer 14

you always multiply by x/x when the second equation is a fraction that has x?

Answer 15

If you want to combine a number with a fraction that has denominator x, then yes.

Answer 16

If the fraction had denominator 2x, then you would multiply the number by 2x/2x

Answer 17

How would you add
\[6 + \frac{6}{7}\]

Answer 18

so something like (3-x) - (3/6x) would mean you multiply by 6x/6x?

Answer 19

you would multiply (3-x) by 6x/6x, yes correct

Answer 20

you multiply the 6 by 7/7?

Answer 21

wouldn't it give an improper fraction?

Answer 22

You would get the following:

\[\frac{7}{7} \times \frac{6}{1} = \frac{42}{7}\]

Answer 23

Then you would add that to 6/7 to get:

\[\frac{42}{7} + \frac{6}{7} = \frac{48}{7}\]

Answer 24

But yes, 42/7 is an improper fraction. Notice that 42/7 = 6

But putting 6 in the proper fraction form with common denominator 7 is the only way to add it to fraction 6/7

Answer 25

Got it. Can you help me with the this one: g/f(x)

Answer 26

(g/f)(x) = g(x)/f(x)

Answer 27

That's the general formula for dividing functions.

Answer 28

1/x / 2-x?

Answer 29

Yes, but to make it easier on yourself, write it like this:
\[1/x \div 2 - x\]

Answer 30

This way, you automatically know to inverse 2 - x

Answer 31

\[\frac{1}{x} \times \frac{1}{2-x}\]

Answer 32

Why is it multiplication now?

Answer 33

Suppose we just had numbers:



Then what would you do?

Answer 34

multiply the 3 by 2/2

Answer 35

We're dividing not adding, so different rules apply. You should go back to learning about taking reciprocals of fractions when you're dividing by them.

Answer 36

but would it be 1/2x-x^2