Thread: Easy Questions, need a reminder

1. Easy Questions, need a reminder

How will I do that for:

f (x) = x^2 - 3 g (x) = 2 - square root (x+1)

Basically the way i did it was.

Add: x^2 - 3 + 2 - square root of (x+1)
= x^2 - square root of (x+1) - 1

Subtract: x^2 - 3 - 2 + square root (x+1)
= x^2 + square root (x+1) - 5

Multiply: I'm not sure how to do this one ... but i got:

f (x) . g (x) = (x^2-3) (2- square root (x+1)

= 2x^2 - square root (x+1)^3 - (square root (x+1)^3 + 3 square root (x+1)

but it's probably wrong.

Can someone help me with these, thanks.

Divide: i'll figure it out after i know how to multiply

2. I'm not sure where you got that...

$\displaystyle (x^2-3) (2- \sqrt{x+1}) = 2x^2 - x^2\sqrt{x+1} - 6 + 3\sqrt{x+1}$

3. Actually for the 3rd part: multiplication I got:

(x^2-3) X (2- square root (x+1)

= 2x^2 - square root (x+1)^3 - 6 + 3 square root (x+1)

= square root (x+1)^3 + 2x^2 + 3 square root (x+1) - 6

Thanks I think I sort of get it. Did I do the adding/subtracting right?

4. $\displaystyle x^2 \times \sqrt{x+1} = x^2\sqrt{x+1} = \sqrt{x^4(x+1)} = \sqrt{x^5 + x^4}$

I don't know how you got $\displaystyle \sqrt{(x+1)^3}$

5. Thanks, I made an error when I tried to multiply it... I havne't doen math in a while and forget a lot of the more simpler stuff... I understand all the basics of rational functions but I find it harder when adding this stuff up sometimes. If I had gotten:

2x^2 - x^2 square root (x+1) - 6 + x^2 square root (x+1)

how would I add it then? I made those numbers up.... but how would i add x^2 quare root (x+1) with x^2 square root (x+1) ... would it just become x^4 square root (x+1) ?

and if i had x^2 square root (x+1) + x^3 square root (x+1) ... would i still add the two to become x^5 (square root x+1) or do i leave it since hte exponenets are different?

thanks.

6. It's okay, and it's good that you want to go through them again!

2x^2 - x^2 square root (x+1) - 6 + x^2 square root (x+1)

how would I add it then? I made those numbers up.... but how would i add x^2 quare root (x+1) with x^2 square root (x+1) ... would it just become x^4 square root (x+1) ?
One way for you to be sure, you can introduce something else.

Let $\displaystyle y = x^2\sqrt{x+1}$

Then

$\displaystyle 2x^2 - x^2\sqrt{x+1} - 6 + x^2\sqrt{x+1} = 2x^2 - y - 6 + y$

How would you deal with this? It's simple like that, right?

and if i had x^2 square root (x+1) + x^3 square root (x+1) ... would i still add the two to become x^5 (square root x+1) or do i leave it since hte exponenets are different?
$\displaystyle x^2 \sqrt{x+1} + x^3\sqrt{x+1}$

No no. I see that you are confusing something.

$\displaystyle x^2 + x^3$ it cannot be further simplified (other than by factoring giving $\displaystyle x^2(1+ x)$

$\displaystyle x^2 \times x^3 = x^5$ When you multiply something with the same base (x) but different powers, then you add. If it's division, you subtract.

Here what you can do is only factor $\displaystyle x^2 \sqrt{x+1}$ and simplify:

$\displaystyle x^2 \sqrt{x+1} + x^3\sqrt{x+1} = x^2\sqrt{x+1}(1 + x) = x^2(x+1)^{3/2}$

Ok, I might not reply for a while now; I'm going to bed. See you tomorrow!

7. Thanks for the help, I appreciate it. Can you let me know if I added and subtracted the functions right or not in the original question? If i have then i'll try dividing the 2 myself before I ask any further questions.

8. The reason I didn't tell you about them is that because they're correct