# Thread: 2 questions i need tips with

1. ## 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

img.top {vertical-align:15%;}

$y=4-x^2$ and the line

img.top {vertical-align:15%;}

$y=3x$
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

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

How would i simplify this?

2. 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?
$2x + 3x^2$

3. Thanks

4. Originally Posted by colby2152
$2x + 4x^2$
Actually $x^{1/2}(2x^{1/2} + 3x^{3/2}) = 2x + 3x^2$

-Dan

5. Originally Posted by Hydralore
Code:
Find the line enclosed between the curve

img.top {vertical-align:15%;}

$y=4-x^2$ and the line

img.top {vertical-align:15%;}

$y=3x$
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

6. Originally Posted by Hydralore
edit : unfortunetly i have another tiny small problem

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

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

7. 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?

8. 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.

9. 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?

10. 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

$\int_{ - 4}^1 {\Big[ {f(x) - g(x)} \Big]\,dx} = \int_{ - 4}^1 {\left( {4 - x^2 - 3x} \right)\,dx} = \frac{{125}}
{6}.$

11. 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.

12. Thanks alot guys

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

$
\int_{0}^p (x-1)^2 dx=3 find P.
$

14. 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.