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\[ ³√1695 - 8x^2\] ?

seems unlikely... missing something?

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

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 \(b=7\)

thanks

\[\lim_{h \rightarrow 0} \frac{ f(13 + h) - f(13) }{ h }\] 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
\[\frac{\sqrt[3]{1695-8(13+h)^2}-7}{h}\]\]

yep

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

university online homework

mathXL

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
\[\frac{-16x}{3(1695-8x^2)^{\frac{2}{3}}}\]

ohh ok thanks!

yeah i think i got the same

yw