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## anonymous 3 years ago 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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1. anonymous

$³√1695 - 8x^2$ ?

2. anonymous

seems unlikely... missing something?

3. anonymous

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

4. anonymous

k

5. 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

6. anonymous

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

7. anonymous

o i see

8. anonymous

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

9. anonymous

thanks

10. anonymous

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

11. anonymous

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

12. anonymous

what is $$f(13)$$?

13. anonymous

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

14. anonymous

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

15. anonymous

yep

16. anonymous

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

17. 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

18. anonymous

university online homework

19. anonymous

mathXL

20. 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

21. anonymous

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

22. anonymous

ill try taking the derivative

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. 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

27. anonymous

ohh ok thanks!

28. anonymous

yeah i think i got the same

29. anonymous

yw

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