Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909

another question from unit 1 - revision sheet - HSN forum

another question from unit 1 - revision sheet

3 replies to this topic

#1xClairex

Top of the Class

• Members
• 296 posts
• Location:west lothian
• Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
• Gender:Female

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?

#2st-and Paul

Top of the Class

• Members
• 356 posts
• Location:St Andrews
• Interests:Check facebook, i cant be bothered putting them all down again :P
• Gender:Male

Posted 15 January 2007 - 10:57 PM

QUOTE(xClairex @ Jan 15 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?

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.

#3xClairex

Top of the Class

• Members
• 296 posts
• Location:west lothian
• Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
• Gender:Female

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

#4xClairex

Top of the Class

• Members
• 296 posts
• Location:west lothian
• Interests:like going out socialising with friends - going to the cinema,bowling, ice skating and shopping :-)<br />like listening to music (listen to all types of music well except classical), and going on msn chatting to friends (oh yes what an exciting life!)
• Gender:Female

Posted 17 January 2007 - 04:42 PM

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

1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users