need help within 12 hours pre calc

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Mar 11th 2009, 06:25 PM
jamman790

need help within 12 hours pre calc

Hey guys i need help with this question using the formula

F( x + h) - F ( x )
-----------------
h

The question being:

F ( x ) = x^1/3

please with a breif explanation of how its done as well:) thank you!
Mar 11th 2009, 06:39 PM
Prove It

Quote:

Originally Posted by jamman790

Hey guys i need help with this question using the formula

F( x + h) - F ( x )
-----------------
h

The question being:

F ( x ) = x^1/3

please with a breif explanation of how its done as well:) thank you!

If $f(x) = x^{\frac{1}{3}}$ then $f(x + h) = (x + h)^{\frac{1}{3}}$ .

You're just substituting $x + h$ where $x$ was.

So

$\frac{f(x + h) - f(x)}{h} = \frac{(x + h)^{\frac{1}{3}}-x^{\frac{1}{3}}}{h}$ .
Mar 11th 2009, 06:50 PM
Plato

Here is the KEY to this question.
$x^3 - y^3 = (x - y )(x^2 + xy + y^2 )$ .
Now, you may not grasp that at first!
But think about.
Mar 11th 2009, 07:18 PM
jamman790

I understand that your subing them in but where do you go from there? i mean how do you expand that?...with the perfect cubes formula listed above?..how does that work? there not being cubed there to the ^1/3..sorry im trying to think about it but am confuzzled:|! Wait do you sub a variable?
Mar 11th 2009, 08:21 PM
Soroban

Hello, jamman790!

We want the Difference Quotient: . $\frac{f(x+h) - f(x)}{h}$

Break it up into three steps:

(1) Find $f(x+h)$ . . . Replace $x$ with $x+h$ ... and simplify.

(2) Subtract $f(x)$ . . . Subtract the original function ... and simplify.

(3) Divide by $h$ . . . Reduce and simplify.

Quote:

$f(x) \:=\:x^{\frac{1}{3}}$

(1) Find $f(x+h)\!:\quad f(x+h) \:=\:(x+h)^{\frac{1}{3}}$

(2) Subtract $f(x)\!:\quad f(x+h)-f(x) \;=\;(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}$

. . .Multiply top and bottom by: . $(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + h^{\frac{2}{3}}$

. . . $\frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{1}\cdot \frac{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}x^{\frac{1}{3}} + h^{\frac{2}{3}}} {(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}h^{\frac{1}{3}} + h^{\frac{2}{3}}}$

. . . $= \;\frac{(x+h) - x}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}h^{\frac{1}{3}} + h^{\frac{2}{3}} }$

. . . $\;=\;\frac{h}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}h^{\frac{1}{3}} + h^{\frac{2}{3}}}$

(3) Divide by $h$

. . . $\frac{f(x+h)-f(x)}{h} \;=\;\frac{{\color{red}\rlap{/}}h}{{\color{red}\rlap{/}}h\left[(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}h^{\frac{1}{3}} + h^{\frac{2}{3}}\right]}$ $=\;\frac{1}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}h^{\frac{1}{3}} + h^{\frac{2}{3}}}$
Mar 11th 2009, 11:23 PM
jamman790

when you multiply the top and bottom by that long equation..how did u figure out you needed to multiply it by that?...