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!

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- Mar 11th 2009, 06:25 PMjamman790need 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 PMProve It
If $\displaystyle f(x) = x^{\frac{1}{3}}$ then $\displaystyle f(x + h) = (x + h)^{\frac{1}{3}}$.

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

So

$\displaystyle \frac{f(x + h) - f(x)}{h} = \frac{(x + h)^{\frac{1}{3}}-x^{\frac{1}{3}}}{h}$. - Mar 11th 2009, 06:50 PMPlato
Here is the KEY to this question.

$\displaystyle x^3 - y^3 = (x - y )(x^2 + xy + y^2 )$.

Now, you may not grasp that at first!

But think about. - Mar 11th 2009, 07:18 PMjamman790
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 PMSoroban
Hello, jamman790!

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

Break it up into three steps:

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

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

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

Quote:

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

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

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

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

. . .$\displaystyle \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}}} $

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

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

(3) Divide by $\displaystyle h$

. . .$\displaystyle \frac{f(x+h)-f(x)}{h} \;=\;\frac{{\color{red}\rlap{/}}h}{{\color{red}\rlap{/}}h\left[(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}h^{\frac{1}{3}} + h^{\frac{2}{3}}\right]}$ $\displaystyle =\;\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 PMjamman790
when you multiply the top and bottom by that long equation..how did u figure out you needed to multiply it by that?...