- anonymous

find the limit as h approaches 0:
f(13+h) - f(13) divided by h
if f(x) = ³√1695 - 8x^2

- jamiebookeater

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- anonymous

\[ ³√1695 - 8x^2\] ?

- anonymous

seems unlikely... missing something?

- anonymous

\[\lim_{h \rightarrow 0}\frac{ \sqrt[3]{1695 - 8x^2} -7 }{ h } \]

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## More answers

- anonymous

k

- anonymous

same gimmick as with a square root, rationalize the numerator, but this time instead of using
\[(a-b)(a+b)=a^2-b^2\] you have to use
\[(a-b)(a^2+ab+b^2)=a^3-b^3\] so it is a pain in the arse

- anonymous

the answer is \[\frac{ -208 }{ 147 }\] i want to figure what to do to get it.

- anonymous

o i see

- anonymous

have fun
but you can do it
use \(a=\sqrt[3]{1695-8x^2}\) and \(b=7\)

- anonymous

thanks

- anonymous

\[\lim_{h \rightarrow 0} \frac{ f(13 + h) - f(13) }{ h }\] this is the original question btw

- anonymous

if f(x) is \[\sqrt[3]{1695 - 8x^2}\]

- anonymous

what is \(f(13)\)?

- anonymous

\[\sqrt[3]{1695 - 8x^2} \] with 13 plugged in, which equals 7

- anonymous

ok then you can start with
\[\frac{\sqrt[3]{1695-8(13+h)^2}-7}{h}\]\]

- anonymous

yep

- anonymous

you can leave it in this form, or you can write
\[\sqrt[3]{343-h^2-208h}-7\]

- anonymous

who gave you this problem? this really sucks
unless you are supposed to use a shortcut, namely recognize this as the derivative and evaluate

- anonymous

university online homework

- anonymous

mathXL

- anonymous

do you know how to take a derivative? because then it it would be not so hard
but if you do not, then there is a ton of work to be done

- anonymous

i do know, im not sure if it would give me the answer im supposed to get

- anonymous

ill try taking the derivative

- anonymous

this is the derivative of
\[\sqrt[3]{1695-8x^2}\] evaluated at \(x=13\)

- anonymous

so if you can take the derivative, then plug in 13, you will get your answer

- anonymous

just to finish quick derivative is
\[\frac{-16x}{3(1695-8x^2)^{\frac{2}{3}}}\]

- anonymous

by the chain rule and power rule
replace \(x\) by 13 and you should get your answer
this is a much snappier way then doing it by hand

- anonymous

ohh ok thanks!

- anonymous

yeah i think i got the same

- anonymous

yw

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