math doubts [NEW]

[Ghost Of Highbury]

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

[nid404]

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 

[Ghost Of Highbury]

it says express it in the form of a - (x-b)^2

 9 - (x-3)^2

therefore a = 9
             b = 3

[nid404]

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

Ur right

[Ghost Of Highbury]

yah...try the mininimum point wala q..i'm not gettingd answer for that....

[nid404]

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]

slvri has answered it correctly....it says find the values of x for which f(x) is greater than 15

[slvri]

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?

[nid404]

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

[Ghost Of Highbury]

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."

[slvri]

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

[Ghost Of Highbury]

can we find out the maximum point by differentiation?

[slvri]

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)

[nid404]

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

[slvri]

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