iop360
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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Algebraic!
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\[ ³√1695 - 8x^2\] ?
Algebraic!
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seems unlikely... missing something?
iop360
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\[\lim_{h \rightarrow 0}\frac{ \sqrt[3]{1695 - 8x^2} -7 }{ h } \]
Algebraic!
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k
anonymous
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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
iop360
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the answer is \[\frac{ -208 }{ 147 }\] i want to figure what to do to get it.
iop360
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o i see
anonymous
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have fun
but you can do it
use \(a=\sqrt[3]{1695-8x^2}\) and \(b=7\)
iop360
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thanks
iop360
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\[\lim_{h \rightarrow 0} \frac{ f(13 + h) - f(13) }{ h }\] this is the original question btw
iop360
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if f(x) is \[\sqrt[3]{1695 - 8x^2}\]
anonymous
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what is \(f(13)\)?
iop360
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\[\sqrt[3]{1695 - 8x^2} \] with 13 plugged in, which equals 7
anonymous
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ok then you can start with
\[\frac{\sqrt[3]{1695-8(13+h)^2}-7}{h}\]\]
iop360
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yep
anonymous
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you can leave it in this form, or you can write
\[\sqrt[3]{343-h^2-208h}-7\]
anonymous
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who gave you this problem? this really sucks
unless you are supposed to use a shortcut, namely recognize this as the derivative and evaluate
iop360
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university online homework
iop360
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mathXL
anonymous
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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
iop360
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i do know, im not sure if it would give me the answer im supposed to get
iop360
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ill try taking the derivative
anonymous
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this is the derivative of
\[\sqrt[3]{1695-8x^2}\] evaluated at \(x=13\)
anonymous
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so if you can take the derivative, then plug in 13, you will get your answer
anonymous
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just to finish quick derivative is
\[\frac{-16x}{3(1695-8x^2)^{\frac{2}{3}}}\]
anonymous
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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
iop360
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ohh ok thanks!
iop360
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yeah i think i got the same
anonymous
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yw