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math doubts [NEW]

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zara:
another question

The function h is defined by

h:x > 6x-x2 for x>=3

Express 6x-x2 in the form a-(x-b)2, where a and b are positive constants.

zara:
3) 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.


i did the expression part...buh im confused in the second part...how to show it? :-\
n yea one more thing....what does this mean "x E R"?
i don't remember our sir explaining this to us....n its der in our worksheet!

Ghost Of Highbury:
ok as he is not online...i wud like to explain it to u..

the first one...u multiply it by x^4 so that u can simplify it and solve easily...as the divisor can be cancelled..

wen u multiply x^4 with 18/x^4 + 1/x^2 = u get => 18 (as the x^4 gets cancelled) + x^4/x^2 = 18 + x^2

next...

(4x^2 - 9)(x^2 + 2) = 0

this means that either (4x^2 - 9) = 0...or (x^2 +2) = 0....as something HAS to be multiplied by 0 to get a 0.

so 4x^2 - 9 = 0

4x^2 = 9..

and so on...

hope that helped :)

Ghost Of Highbury:
x E R means that the values of x belong to the set of real values...


PS: wud try to solve thos.e

zara:
thanks adi!!
tht helped!
n yea sure go ahead solve it!
its nt for specific people...whoever can is free to do! =)

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