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Past Paper Help <3 2002 winter diet - polynomials, functions

Sammie

Posted 09 February 2008 - 04:14 PM

Polynomals - Non Calc Q -5

Given that (x-2) and (x+3) are factors of f(x) = 3x3+2x2+cx+d find the values of c and d.

working so far:

2|3 2 c d
6 16 16+2c
______________
3 8 16+c 16+2c+d

-3|3 2 c d
-9 21 -63-3c
________________
3 -7 21+c -63-3c+d

Using simultaimous eq.

32+2c+d =0
and
-63-3c+d=0

Therefore:

-32=2c+d
63=-3c+d

thats where I got stuck...can't really see where I've messed up in the question and I don't really know where to go from there

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Functions - 2002 Winter Diet Q - 9

a] The function f, is defined on a suitable domain, is given by f(x) = 3/(x+1) find an expression for h(x) = (f(fx)) giving your answer as a fraction in the simplest form.

working so far

f(x) = 3/(x+1)
h(x) = f(f(x))
h(x) = 3/(3/x+1)+1

I worked it though every way I could think of and even though the answers say 3(x+1)/(x+4) I don't see how on earth to get that.

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Functions - 2002 Winter Diet Q - 10

A function f is defined by f(x) = 2x+3+(18/(x-4) where x is no equal to 4. Find the values of x for which the funtion is increasing.

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Thank to anyone who tries to help, I know its a lot but my prelims in like 2 weeks and I just go tmy uni offers though and I wanna try and avoid the screams of help echoing though the forum that I caused last year.

xo

Arisa

Posted 09 February 2008 - 08:42 PM

Polynomials:

I think you are doing right so far, just need to solve the simultaimous equations:

-30=2c+d---eq.1
63=-3c+d---eq.2

eq.1 x 3: -96=6c+3d
eq.2 x 2: 126=-6c+2d

eq.1+eq.2: 30=5d

so, d=6, and if you substitute it into eq.1, you get c=-19

Functions(Q9):

f(f(x))=3/((3/(x+1))+1)

Multplying (x+1) to the top and the bottom:
f(f(x))=3/((3/(x+1))+1) x (x+1)/(x+1)
f(f(x))=3(x+1)/(3+1(x+1))

Therefore
f(f(x))=3(x+1)/(x+4)

Do you see what I've done? I think the most important bit is to multply (x+1) to the top and the bottom to avoid triple decker fraction(if you know what I mean....)

Functions(Q10):

If the question says " increasing", "maximum" etc, the best method you could use is DIFFERENTIATION (the most exciting bit of maths biggrin.gif ).

f(x)=2x+3+18/(x-4)=2x+3+18(x-4)^-1
so f'(x)=2-18(x-4)^-2 = 2-18/(x-4)^2

f'(x)=0 when the gradiant is 0,
2-18/(x-4)^2=0
2(x-4)^2-18=0
2(x-4)^2=18
(x-4)^2=9
(x-4)=+3 or -3(a square root of 9 is plus or minus 3)
so x=7 or 1
Since f(x) is increasing, f'(x)>0 so the answer is x<1 or x>7

Hopefully it helps you, and good luck for your prelim!!! smile.gif

Sammie

Posted 10 February 2008 - 01:27 PM

Thank you very much thats awesome. Its easy enough when I look at it now, its just knowing what to use where <3 can't believe I didn't get the first one now. Thanks again, I'm gonna ace this prelim if it kills me <3 xo

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