Question

Operation with functions. f+g, f-g, fg,f/g

a.f(x) = ^x-2, g(x)=3x

b. F(x) = 1/x-2; g(x)= 1/x

Answer 1

Hey,, I'm confused can you add or subtract if the given is square root?

Answer 2

You can't add square roots to other things, generally.

Answer 3

And do I have to have the same denominator if I'm dealing with fraction in operations?

Answer 4

how bout subtracting or multiply?

Answer 5

I can't really answer your question specifically because I don't know what your functions are. What does f(x) = ^x-2 and F(x) = 1/x-2 mean?

Answer 6

@ParthKohli I can't message you :/

Answer 7

ohh wait

Answer 8

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

Answer 9

\[g(x) 3x\]

Answer 10

So basically I have to add this function. But I'm having doubt if I can add if the function is rooted.

Answer 11

Sadly, you cannot. It is simply \(\sqrt{x-2}+3x\).

Answer 12

So what is the Domain and Range of the new Function?

Answer 13

Depends if you're working in \(\mathbb{R}\) or \(\mathbb{C}\). :P

Most times, it is \(\mathbb{R}\). The square root function is not defined in \(\mathbb{R}\) for all input less than \(0\).

So, set the inside of your square root to \(0\) and solve for \(x\).

Answer 14

I'm confused. What's R and C?

Answer 15

The real numbers and the complex numbers.

Answer 16

Correct me if I'm wrong

Answer 17

f+g = \[\sqrt{x-2} + 3x \]
\[f-g = \sqrt{x-2} -3x\]
\[fg = ( \sqrt{x-2} ) 3x\]
\[f/g = \sqrt{x-2} / 3x\]

Answer 18

So I can't multiply them if it's rooted?

Answer 19

@Limitless You still there?

Answer 20

@ParthKohli Help I'm confused :<

Answer 21

Yes

Answer 22

You have it right.

Answer 23

*sighs* So how about the fractions? Do i have to make the denominators the same?

Answer 24

No need to modify what you've done so far. All your expressions are correct. :) However, if you wish to find the domain and range, that's a different question.

Answer 25

Some people prefer \(\frac{f}{g}=\frac{\sqrt{x-2}}{3x}\), but that is a cosmetic detail.

Answer 26

Okay, how about F(x) = 1/x-2; g(x)= 1/x?

Answer 27

They have different denominators so I can't add them

Answer 28

Can you put F(x) in latex notation?

Answer 29

Okay

Answer 30

\[f(x) = 1 \div x-2\]

Answer 31

\[g(x) = 1\div x\]

Answer 32

Okay. To add them, make them have the same denominator:
\[\frac{1}{x-2}+\frac{1}{x}=\frac{x}{x(x-2)}+\frac{(x-2)}{(x-2)x}=\frac{x+(x-2)}{x(x-2)}=\frac{2x-2}{x(x-2)}\]
Some people prefer the denominator simplified, though. So,
\[\frac{f}{g}=\frac{2x-2}{x^2-2x}\]

Answer 33

So the rule in adding fraction is still applied? I thought Since they're Function you can't do that.

Answer 34

Oh my lord, I need sleep. I meant: \[f+g=\frac{2x-2}{x^2-2x}\]

Answer 35

Terribly sorry for that.

Answer 36

NOw I get it . Now I"m only concerned of division.

Answer 37

\[\frac{f}{g}=\frac{\frac{1}{x-2}}{\frac{1}{x}}=\frac{\frac{1}{x-2}}{\frac{1}{x}} \cdot\frac{x}{x}=\frac{\frac{x}{x-2}}{1}=\frac{x}{x-2}\]
:D

Answer 38

:D So you do cancellation. If I multiply fractions with different denominators. Do I also need to find the LCD ?

Answer 39

Also what is the reasoning behing the multiplication of x to the numerators? ^

Answer 40

Not really. You just try to find the most simple form possible. You multiply the numerators and denominators and figure out what comes out. :D

I multiplied by \(\frac{x}{x}\) because it cancelled the denominator and left us with only one fraction. (You can multiply by \(\frac{x}{x}\) because it's \(1\).)

Answer 41

Okay. Thanks alot!

Answer 42

You're welcome. Goodluck on future maths.