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Higher Maths problem

How would do you get the answer for this, i am completely stuck, please help!

6. Express x(powerof2) + 2x + 9 in the form (x + a)powerof2 + b and hence state the maximum value of

16
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x(powerof2) + 2x + 9

The line is so supposed to be divided. Sorry I can do powers!

Thanks in advance
Reply 1
Avatar for ProudScottish
ProudScottish
OP
9
Pheylan
start by completing the square on x2+2x+9x^2+2x+9

Sorry but the thing I don't have a clue how to complete the square and I didn't get even when the teacher told me and he got fed up and just went away to someone else :rolleyes: !
ProudScottish
Sorry but the thing I don't have a clue how to complete the square and I didn't get even when the teacher told me and he got fed up and just went away to someone else :rolleyes: !


I'll say upfront that you probably want to just remember the equation above, or at least that's what I did. But in terms of deriving it:

You want to place a polynomial of the form x2+bx+cx^2 + bx + c in the form (x−p)2+q(x-p)^2 + q. So expand out the right hand side (the form you're aiming for) to get x2−2px+p2+qx^2 - 2px + p^2 + q. Then just equate coefficients: −2p=b-2p = b, p2+q=cp^2 + q = c.

So, substituting back in, you've got (x+b2)2+c−(b2)2\left(x + \tfrac{b}{2}\right)^2 + c - \left(\tfrac{b}{2}\right)^2. If the leading coefficient isn't 1 (e.g. 3x2+2x+13x^2 + 2x + 1) then you just divide through by it and proceed as before.
Reply 3
Avatar for Armaros
Armaros
6
TheUnbeliever
I'll say upfront that you probably want to just remember the equation above, or at least that's what I did. But in terms of deriving it:

You want to place a polynomial of the form x2+bx+cx^2 + bx + c in the form (x−p)2+q(x-p)^2 + q. So expand out the right hand side (the form you're aiming for) to get x2−2px+p2+qx^2 - 2px + p^2 + q. Then just equate coefficients: −2p=b-2p = b, p2+q=cp^2 + q = c.

So, substituting back in, you've got (x+b2)2+c−(b2)2\left(x + \tfrac{b}{2}\right)^2 + c - \left(\tfrac{b}{2}\right)^2. If the leading coefficient isn't 1 (e.g. 3x2+2x+13x^2 + 2x + 1) then you just divide through by it and proceed as before.

Sorry to hijack this thread, but I have my own little problem, and I don't think it's really worth its own:

if f(x)=1−xxf(x) = \tfrac{1-x}{x} then what is f(f(x))f(f(x)) in its simplest form?

I get to 1−1−xx1−xx\dfrac{1-\tfrac{1-x}{x}}{\tfrac{1-x}{x}} and then just fail. Can anybody help me continue, or tell me what I've already done wrong? Thank you~
Armaros
Sorry to hijack this thread, but I have my own little problem, and I don't think it's really worth its own:

if f(x)=1−xxf(x) = \tfrac{1-x}{x} then what is f(f(x))f(f(x)) in its simplest form?

I get to 1−1−xx1−xx\dfrac{1-\tfrac{1-x}{x}}{\tfrac{1-x}{x}} and then just fail. Can anybody help me continue, or tell me what I've already done wrong? Thank you~


Multiply top and bottom by x:

x−(1−x)1−x=2x−11−x\displaystyle\frac{x-(1-x)}{1-x} = \frac{2x-1}{1-x}

Polynomial division (or 'by inspection' in this simple case) shows 2x−11−x=−2+11−x\frac{2x-1}{1-x} = -2 + \frac{1}{1-x}. Whether that's actually any more simple is debatable; I'd be tempted to leave it as a single fraction.
Reply 5
Avatar for Armaros
Armaros
6
TheUnbeliever
Multiply top and bottom by x:

x−(1−x)1−x=2x−11−x\displaystyle\frac{x-(1-x)}{1-x} = \frac{2x-1}{1-x}

Polynomial division (or 'by inspection' in this simple case) shows 2x−11−x=−2+11−x\frac{2x-1}{1-x} = -2 + \frac{1}{1-x}. Whether that's actually any more simple is debatable; I'd be tempted to leave it as a single fraction.

Oh, good gracious, I can't believe how I didn't see that. Thanks a lot!

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