Author Topic: math doubts [NEW]  (Read 16157 times)

Offline Ghost Of Highbury

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Re: functions
« Reply #30 on: September 27, 2009, 07:20:05 am »
the question says....."Express 6x-x2 in the form a-(x-b)2"...

if the answer is (x+3)2 -9 ...it simplifies to x2 + 6x + 9 -9 = x2 + 6x

but the question says that it shud be -x2 + 6x

for this the answer should be...

9 - (x-3)2 which simplifies to 9 - (x2 - 6x + 9)

==> 9 - x2 + 6x - 9 = -x2 + 6x

???
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nid404

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Re: functions
« Reply #31 on: September 27, 2009, 07:34:54 am »
the question says....."Express 6x-x2 in the form a-(x-b)2"...

if the answer is (x+3)2 -9 ...it simplifies to x2 + 6x + 9 -9 = x2 + 6x

but the question says that it shud be -x2 + 6x

for this the answer should be...

9 - (x-3)2 which simplifies to 9 - (x2 - 6x + 9)

==> 9 - x2 + 6x - 9 = -x2 + 6x

???

yeah right.....the question says 6x-x2...
yup ur right...infact it says express it in the form c-(a+b)2 

Offline Ghost Of Highbury

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Re: functions
« Reply #32 on: September 27, 2009, 07:38:06 am »
it says express it in the form of a - (x-b)^2

 9 - (x-3)^2

therefore a = 9
             b = 3
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nid404

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Re: functions
« Reply #33 on: September 27, 2009, 07:38:58 am »
it says express it in the form of a - (x-b)^2

 9 - (x-3)^2

therefore a = 9
             b = 3

whatever...it just says express it in that form :P

Ur right

Offline Ghost Of Highbury

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Re: functions
« Reply #34 on: September 27, 2009, 07:48:09 am »
yah...try the mininimum point wala q..i'm not gettingd answer for that....
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nid404

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Re: functions
« Reply #35 on: September 27, 2009, 07:59:38 am »
yah...try the mininimum point wala q..i'm not gettingd answer for that....

There r too many questions and answers...I'm getting confused....can you post the question again....that will make it easier

nid404

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Re: functions
« Reply #36 on: September 27, 2009, 08:07:27 am »
slvri has answered it correctly....it says find the values of x for which f(x) is greater than 15

Offline slvri

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Re: functions
« Reply #37 on: September 27, 2009, 08:12:07 am »
slvri has answered it correctly....it says find the values of x for which f(x) is greater than 15
hey nid.......need any other q's solved?
i hate A level...........

nid404

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Re: functions
« Reply #38 on: September 27, 2009, 08:13:48 am »
hey nid.......need any other q's solved?

Well not for now....actually adi did not get the the domain for the f(x)>15 thingy....so I was trying to explain to him why you were correct...hope he got it

Offline Ghost Of Highbury

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Re: functions
« Reply #39 on: September 27, 2009, 08:14:35 am »
ya slvri...this one...this q wasnt answered...

" f is defined by:  f:x > 3x-2 for x E R

   g is defined by: g:x >6x-x2 for x E R

Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9."
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Offline slvri

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Re: functions
« Reply #40 on: September 27, 2009, 08:25:13 am »
ya slvri...this one...this q wasnt answered...

" f is defined by:  f:x > 3x-2 for x E R

   g is defined by: g:x >6x-x2 for x E R

Express gf(x) in terms of x, and hence show that the maximum value of gf(x) is 9."
gf(x)=g(f(x))=g(3x-2)
=6(3x-2)-(3x-2)2
=18x-12-(9x2-12x+4)
=18x-12-9x2+12x-4
=-16+30x-9x2
=-16-(-30x+9x2)
=-16-(9x2-30x)
=-16-9(x2-(10/3)x)
=-16-9((x)2-2(x)(5/3)+(5/3)2)+9(5/3)2
=-16+9(5/3)2-9(x-5/3)2
=-16+25-9(x-5/3)2
=9-9(x-5/3)2
now u can see that when x=5/3
gf(5/3)=9-(5/3-5/3)2=9-0=9
and whatever value u put for x, the value of gf(x) will always be less than or equal to 9
so the maximum value of gf(x)=9
shown
i hate A level...........

Offline Ghost Of Highbury

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Re: functions
« Reply #41 on: September 27, 2009, 08:29:11 am »
can we find out the maximum point by differentiation?
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Offline slvri

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Re: functions
« Reply #42 on: September 27, 2009, 08:40:29 am »
can we find out the maximum point by differentiation?
yes u can assuming this is an a level math or an igcse add math q(but not if this is an igcse math q)
i hate A level...........

nid404

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Re: functions
« Reply #43 on: September 27, 2009, 08:56:50 am »
differentiating becomes much easier

gf(x)=16+30x-9x2

when u differentiate
-18x2+30
x=-5/3

substitute in the equation you get max value of gf(x) as 9

Offline slvri

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Re: functions
« Reply #44 on: September 27, 2009, 09:03:25 am »
differentiating becomes much easier

gf(x)=16+30x-9x2

when u differentiate
-18x2+30
x=-5/3

substitute in the equation you get max value of gf(x) as 9

whatever way u like
i hate A level...........