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