1. ## Precalculus help

I think i know this, but im not exactly sure....

1. simplify:

(x^(-2)-y^(-2))(y+x)^(-1)

2. If f(x) = (4/x), find and simplify (f(x+h)-f(x))/(h)

3. express the area A of an equilateral triangle as a function of the length S of a side of the triangle.

4. solve for the real number x: ln x - ln(x-1) = 1

5. solve for x in the interval [0, 2(pi)]: sin2x-sinx = 0

thanks!!

2. Originally Posted by skabani
I think i know this, but im not exactly sure...

simplify:

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

............... $= \frac {\frac {y^2 - x^2}{x^2y^2}}{y + x}$

............... $= \frac {(y + x)(y - x)}{x^2 y^2} \cdot \frac {1}{y + x}$

............... $= \frac {y - x}{x^2 y^2}$

doesn't look much simpler to me, but whatever.

EDIT: wow, where'd all those other questions come from. anyway, i think i'm going to leave now. you'll have to wait on someone else to help

3. Originally Posted by skabani
2. If f(x) = (4/x), find and simplify (f(x+h)-f(x))/(h)
Aside from a bit of work, this one is straightforward:
$f(x) = \frac{4}{x}$

$f(x + h) = \frac{4}{x + h}$

So
$\frac{f(x + h) - f(x)}{h} = \frac{ \frac{4}{x + h} - \frac{4}{x} }{h}$

$= \frac{ \frac{4x}{x(x + h)} - \frac{4(x + h)}{x(x + h)} }{h}$

$= \frac{ \frac{4x - 4(x + h)}{x(x + h)} } {h}$

$= \frac{4x - 4x - 4h}{hx(x + h)}$

$= \frac{-4h}{hx(x + h)}$

$= -\frac{4}{x(x + h)}$

-Dan

4. Originally Posted by skabani
4. solve for the real number x: ln x - ln(x-1) + 1
You can't solve this. It isn't an equation. You could simplify it though...

$ln(x) - ln(x - 1) + 1 = ln(x) - ln(x - 1) + ln(e) = ln \left ( \frac{ex}{x - 1} \right )$

Originally Posted by skabani
5. solve for x in the interval [0, 2(pi)]: sin2x-sinx = 0
$sin(2x) = 2sin(x)cos(x)$

so...
$sin(2x) - sin(x) = 0$

$2sin(x)cos(x) - sin(x) = 0$

$sin(x) ( 2cos(x) - 1 ) = 0$

So either
$sin(x) = 0 \implies x = 0, ~ \pi$
or
$2cos(x) - 1 = 0 \implies cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, ~ \frac{5\pi}{3}$

-Dan

5. Originally Posted by topsquark
You can't solve this. It isn't an equation. You could simplify it though...

$ln(x) - ln(x - 1) + 1 = ln(x) - ln(x - 1) + ln(e) = ln \left ( \frac{ex}{x - 1} \right )$

$sin(2x) = 2sin(x)cos(x)$

so...
$sin(2x) - sin(x) = 0$

$2sin(x)cos(x) - sin(x) = 0$

$sin(x) ( 2cos(x) - 1 ) = 0$

So either
$sin(x) = 0 \implies x = 0, ~ \pi$
or
$2cos(x) - 1 = 0 \implies cos(x) = \frac{1}{2} \implies x = \frac{\pi}{3}, ~ \frac{5\pi}{3}$

-Dan

thanks..and number 4, was not + 1, it was =1...sorry abt that..

6. Originally Posted by skabani
4. solve for the real number x: ln x - ln(x-1) = 1
Same idea as before:
$ln(x) - ln(x - 1) = 1$

$ln \left ( \frac{x}{x - 1} \right ) = 1$

$\frac{x}{x - 1} = e$

$x = e(x - 1)$

$x = ex - e$

$x - ex = -e$

$x(1 - e) = -e$

$x = \frac{-e}{1 - e} = \frac{e}{e - 1}$

-Dan

7. Originally Posted by Jhevon
$\frac {x^{-2} - y^{-2}}{y + x} = \frac {\frac {1}{x^2} - \frac {1}{y^2}}{y + x}$

............... $= \frac {\frac {y^2 - x^2}{x^2y^2}}{y + x}$

............... $= \frac {(y + x)(y - x)}{x^2 y^2} \cdot \frac {1}{y + x}$

............... $= \frac {y - x}{x^2 y^2}$
Another solution

\begin{aligned}\frac{x^{-2}-y^{-2}}{y+x}&=\frac{x^2y^2\left(x^{-2}-y^{-2}\right)}{x^2y^2(y+x)}\\&=\frac{y^2-x^2}{x^2y^2(y+x)}\\&=\frac{y-x}{x^2y^2}\end{aligned}

8. Originally Posted by topsquark
Same idea as before:
$ln(x) - ln(x - 1) = 1$

$ln \left ( \frac{x}{x - 1} \right ) = 1$

$\frac{x}{x - 1} = e$

$x = e(x - 1)$

$x = ex - e$

$x - ex = -e$

$x(1 - e) = -e$

$x = \frac{-e}{1 - e} = \frac{e}{e - 1}$
Other solution

$\ln x-\ln(x-1)=1\implies\ln\frac{x-1}x=-1\implies\frac1e=1-\frac1x\,\therefore\,x=\frac e{e-1}$