2 questions i need tips with

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December 10th 2007, 02:04 PM
Hydralore

2 questions i need tips with

$ x^\frac{1}{2}(2x^\frac{1}{2}+3x^\frac{3}{2}) $

How would this multiply out so it can be differentiated?

also

Code:

Find the line enclosed between the curve and the line

Iv found the two factors are x+1 and x-4 but does this mean the limit is x+1 and x-4 or x-1 and x+4?

Thanks in advance ;)

edit : unfortunetly i have another tiny small problem :D

$\frac {1}{9x^2+6x+2}$

How would i simplify this?
December 10th 2007, 06:09 PM
colby2152

Quote:

Originally Posted by Hydralore

$ x^\frac{1}{2}(2x^\frac{1}{2}+3x^\frac{3}{2}) $

How would this multiply out so it can be differentiated?

$2x + 3x^2$
December 11th 2007, 07:22 AM
Hydralore

Thanks
December 11th 2007, 07:24 AM
topsquark

Quote:

Originally Posted by colby2152

$2x + 4x^2$

Actually $x^{1/2}(2x^{1/2} + 3x^{3/2}) = 2x + 3x^2$

-Dan
December 11th 2007, 07:26 AM
topsquark

Quote:

Originally Posted by Hydralore

Code:

Find the line enclosed between the curve and the line

Iv found the two factors are x+1 and x-4 but does this mean the limit is x+1 and x-4 or x-1 and x+4?

"Find the line enclosed" between a curve and a line? Is this written correctly?

-Dan
December 11th 2007, 07:32 AM
topsquark

Quote:

Originally Posted by Hydralore

edit : unfortunetly i have another tiny small problem :D

$\frac {1}{9x^2+6x+2}$

How would i simplify this?

In most cases you wouldn't. The quadratic doesn't factor over the rationals. If you really insist, you could write it as:
$\frac {1}{9x^2+6x+2} = \frac{1}{(3x + 1 + i)(3x + 1 - i)}$
but that's kind of overkill for most problems.

-Dan
December 11th 2007, 07:34 AM
colby2152

Quote:

Originally Posted by topsquark

Actually $x^{1/2}(2x^{1/2} + 3x^{3/2}) = 2x + 3x^2$

-Dan

Haha, looks like I added to the coefficient as well?
December 11th 2007, 01:27 PM
Hydralore

Quote:

Originally Posted by topsquark

"Find the line enclosed" between a curve and a line? Is this written correctly?

-Dan

... no.. im just stupid.. its 'find the area enclosed'
sorry.
December 11th 2007, 01:30 PM
Hydralore

Quote:

Originally Posted by topsquark

In most cases you wouldn't. The quadratic doesn't factor over the rationals. If you really insist, you could write it as:
$\frac {1}{9x^2+6x+2} = \frac{1}{(3x + 1 + i)(3x + 1 - i)}$
but that's kind of overkill for most problems.

-Dan

the original question was
$ f(x)= \frac{1}{x^2-1}$

$g(x) = 3x+1 $
find h(x) when h(x) = f(g(x))

im guessing iv made a mistake?
December 11th 2007, 01:55 PM
Krizalid

Quote:

Originally Posted by Hydralore

Find the line enclosed between the curve $y=4-x^2$ and the line $y=3x$

Get the coordinates where these equations intersect.

Define $f(x)=4-x^2$ & $h(x)=3x,$ now

$4-x^2=3x\implies x^2+3x-4=0\,\therefore\,(x-1)(x+4)=0.$

Now we have to integrate a certain function on the inverval $[-4,1].$ The thing is, in such interval $f(x)\ge g(x),$ so the integral to compute will be

$\int_{ - 4}^1 {\Big[ {f(x) - g(x)} \Big]\,dx} = \int_{ - 4}^1 {\left( {4 - x^2 - 3x} \right)\,dx} = \frac{{125}} {6}.$
December 11th 2007, 02:07 PM
Krizalid

Quote:

Originally Posted by Hydralore

$ f(x)= \frac{1}{x^2-1}$

$g(x) = 3x+1 $
find h(x) when h(x) = f(g(x))

$(f \circ g)(x) = \frac{1} {{(3x + 1)^2 - 1}}.$

That's all.
December 11th 2007, 02:13 PM
Hydralore

Thanks alot guys :D
December 11th 2007, 02:18 PM
Hydralore

i have one last question i need help with ...

$ \int_{0}^p (x-1)^2 dx=3 find P. $
December 11th 2007, 02:21 PM
Krizalid

Start by settin' $u=x-1,$ the integral becomes

$\int_{ - 1}^{p - 1} {u^2 \,du} = \left. {\frac{1} {3}u^3 } \right|_{ - 1}^{p - 1} = \frac{{(p - 1)^3 + 1}} {3} = 3.$

Now you know what to do.