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another question from unit 1 - revision sheet - HSN forum

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another question from unit 1 - revision sheet


3 replies to this topic

#1 xClairex

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Posted 15 January 2007 - 09:37 PM

aah so dont get all this basic stuff

the question is

find f'(4) where f(x)= x-1/ ( root x)


so far i have f(x)= (x to the power of 1/2) - (x to the power of -1/2)

this is worth 5 marks so i think i would then do

f'(x) = (x to the power of -1/2) - (x to the power of -3/2)

then sub in 4

f'(4) = (4 to the pwer of -1/2) - (4 to the power of -3/2)

but this is a non calculator question so i dont know what to do next

can anyone help?


#2 st-and Paul

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Posted 15 January 2007 - 10:57 PM

QUOTE(xClairex @ Jan 15 2007, 09:37 PM) View Post

aah so dont get all this basic stuff

the question is

find f'(4) where f(x)= x-1/ ( root x)


so far i have f(x)= (x to the power of 1/2) - (x to the power of -1/2)

this is worth 5 marks so i think i would then do

f'(x) = (x to the power of -1/2) - (x to the power of -3/2)

then sub in 4

f'(4) = (4 to the pwer of -1/2) - (4 to the power of -3/2)

but this is a non calculator question so i dont know what to do next

can anyone help?


You are right in what you have done so far. You are making things difficult for yourself by trying to calculate the f'(4) using the form of the derivative you have already. Rewrite the derivative as: 1/x^1/2 - 1/x^3/2
This will give 1/2 - 1/8 as (4^3)^1/2 is (64)^1/2.

Therefore your final answer should be 3/8. All i did was use the law of indices to write the negative powers x^-n as 1/x^n

This makes the evaluation part a lot easier.

Hope that helped.


#3 xClairex

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Posted 16 January 2007 - 04:53 PM

Rewrite the derivative as: 1/x^1/2 - 1/x^3/2 i understand this bit

This will give 1/2 - 1/8 as (4^3)^1/2 is (64)^1/2. but it is this bit that im not sure about can you explain how you got this?

Therefore your final answer should be 3/8. All i did was use the law of indices to write the negative powers x^-n as 1/x^n

This makes the evaluation part a lot easier.

thanks for the help it's much appreciated



#4 xClairex

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Posted 17 January 2007 - 04:42 PM

its ok i know how to do this now thanx for the help






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