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

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\[ ³√1695 - 8x^2\] ?
seems unlikely... missing something?
\[\lim_{h \rightarrow 0}\frac{ \sqrt[3]{1695 - 8x^2} -7 }{ h } \]

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Other answers:

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
the answer is \[\frac{ -208 }{ 147 }\] i want to figure what to do to get it.
o i see
have fun but you can do it use \(a=\sqrt[3]{1695-8x^2}\) and \(b=7\)
\[\lim_{h \rightarrow 0} \frac{ f(13 + h) - f(13) }{ h }\] this is the original question btw
if f(x) is \[\sqrt[3]{1695 - 8x^2}\]
what is \(f(13)\)?
\[\sqrt[3]{1695 - 8x^2} \] with 13 plugged in, which equals 7
ok then you can start with \[\frac{\sqrt[3]{1695-8(13+h)^2}-7}{h}\]\]
you can leave it in this form, or you can write \[\sqrt[3]{343-h^2-208h}-7\]
who gave you this problem? this really sucks unless you are supposed to use a shortcut, namely recognize this as the derivative and evaluate
university online homework
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
i do know, im not sure if it would give me the answer im supposed to get
ill try taking the derivative
this is the derivative of \[\sqrt[3]{1695-8x^2}\] evaluated at \(x=13\)
so if you can take the derivative, then plug in 13, you will get your answer
just to finish quick derivative is \[\frac{-16x}{3(1695-8x^2)^{\frac{2}{3}}}\]
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
ohh ok thanks!
yeah i think i got the same

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