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math doubts [NEW]

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slvri:

--- Quote from: eddie_adi619 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."

--- End quote ---
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:

--- Quote from: eddie_adi619 on September 27, 2009, 08:29:11 am ---can we find out the maximum point by differentiation?

--- End quote ---
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:

--- Quote from: nid404 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


--- End quote ---
whatever way u like

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