#1

I need some help with some precalculus thing.....

f(x) = x^2 -2x

find the values for which f(x) > 15

find the range of f.

f(x) = x^2 -2x

find the values for which f(x) > 15

find the range of f.

#2

42.

#3

46?? uh, its a bit of work, but could you type out the steps?? Im a bit of a dunce

#4

x ≥ 5

I think... I failed pre cal and did this in my head.

EDIT:

I think... I failed pre cal and did this in my head.

EDIT:

f(x) will be greater than 15 for any value of x greater than 5.

*Last edited by biga29 at Dec 6, 2011,*

#5

f(x) will be greater than 15 for any value of x greater than 5.

#6

The answer is 200. Which also happens to be the wattage of the amp you need.

#7

x ≥ 5

I think, I failed pre cal and did this in my head...

It's only >5, can't equal

TS is shit at math ...

#8

It's only >5, can't equal

TS is shit at math ...

Oh, ok. Because the question was for everything greater than 15, not greater than or equal to 15 right?

#9

Oh, ok. Because the question was for everything greater than 15, not greater than or equal to 15 right?

Yes, and also TS is shit at math

#10

It's only >5, can't equalTS is shit at math...

I know!

#11

0x

#12

ok, find the coefficient of x^4 in the expansion of (2-x^2)(1-2x^2)^8

#13

ok, find the coefficient of x^4 in the expansion of (2-x^2)(1-2x^2)^8

Ok, find the coefficient of x^4 in the expansion of (2-x^2)(1-2x^2)^8, yourself

#14

This isn't what your mom meant when she said to do your homework.ok, find the coefficient of x^4 in the expansion of (2-x^2)(1-2x^2)^8

Also, there is a huge math thread with a bunch nerds who get boners for stuff like this.

#15

My atttempt at a... I'm a but rusty in the language of Mathematics.

x² – 2x > 15

x² – 2x – 15 > 0

(x – 5)(x + 3) > 0

(x - 5) <- - -(-3)- - -(5)----->

(x + 3) <- - -(-3)-----(5)----->

f(x) .. <-----(-3)- - -(5)----->

- - - Negative

----- Postive

Therefore,

f(x) > 15 when:

x < (-3) or x > 5

When x is either grater than 5 or less than negative 3.

x² – 2x > 15

x² – 2x – 15 > 0

(x – 5)(x + 3) > 0

(x - 5) <- - -(-3)- - -(5)----->

(x + 3) <- - -(-3)-----(5)----->

f(x) .. <-----(-3)- - -(5)----->

- - - Negative

----- Postive

Therefore,

f(x) > 15 when:

x < (-3) or x > 5

When x is either grater than 5 or less than negative 3.

*Last edited by sfaune92 at Dec 6, 2011,*

#16

^ And now I feel rather silly for forgetting that there would also be a negative answer. That's what I get for looking at it and thinking '5' rather than working it through properly...

We're not here to just give you all of your answers.

Expand the brackets. It's simple, although with that power of 8 it might take a couple of minutes. You'll get your coefficient of x^4.

ok, find the coefficient of x^4 in the expansion of (2-x^2)(1-2x^2)^8

We're not here to just give you all of your answers.

Expand the brackets. It's simple, although with that power of 8 it might take a couple of minutes. You'll get your coefficient of x^4.

#17

Expanding the brackets is a really long way of doing the question. Use the binomial expansion to find the coefficients in the second term that, when multiplied by either the 2 or the -x^2 will give x^4.

#18

Just use wolfram like any normal person. http://www.wolframalpha.com/input/?i=x%5E2+-2x%3E15

#19

Just use wolfram like any normal person. http://www.wolframalpha.com/input/?i=x%5E2+-2x%3E15

You should've made him type it in himself

#20

Expanding the brackets is a really long way of doing the question. Use the binomial expansion to find the coefficients in the second term that, when multiplied by either the 2 or the -x^2 will give x^4.

Ah yes, good point. It's been so long since I've done one of those, I think I'd forgotten it existed.

#21

There's a mathematics help thread, use it.

#22

Cheat and use wolframalpha....or you know, learn to do it yourself.