# 2 questions i need tips with

• Dec 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 <br /> img.top {vertical-align:15%;}<br /> $y=4-x^2$ and the line <br /> img.top {vertical-align:15%;}<br /> $y=3x$
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?

edit : unfortunetly i have another tiny small problem :D

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

How would i simplify this?
• Dec 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?

$2x + 3x^2$
• Dec 11th 2007, 07:22 AM
Hydralore
Thanks
• Dec 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
• Dec 11th 2007, 07:26 AM
topsquark
Quote:

Originally Posted by Hydralore
Code:

Find the line enclosed between the curve <br /> img.top {vertical-align:15%;}<br /> $y=4-x^2$ and the line <br /> img.top {vertical-align:15%;}<br /> $y=3x$
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
• Dec 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
• Dec 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?
• Dec 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.
• Dec 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?
• Dec 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

$\int_{ - 4}^1 {\Big[ {f(x) - g(x)} \Big]\,dx} = \int_{ - 4}^1 {\left( {4 - x^2 - 3x} \right)\,dx} = \frac{{125}}
{6}.$
• Dec 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.
• Dec 11th 2007, 02:13 PM
Hydralore
Thanks alot guys :D
• Dec 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.
$
• Dec 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.