Melody

Always a pleasure  :)

Hi EggRick,

How about presenting your questions in the best possible way.

Nobody wants to look at all those irrelevant dollar signs.

Thanks guest, I am just going to look at it a different way :)

Find the coefficient of y^4 in the expansion of (2y - 5 + y^2)^6.

$[(2y - 5) + y^2]^6$

the general term will be      $\binom{6}{r}(y^2)^r(2y-5)^{6-r}$

We are looking for the y^4 term so r can only be 0,1, or 2  (becasue after that I can see the y will have at least a power of 6

you need to do the three seperately and add the x^4 terms together to get the final answer.

I will do r =1,   that is the hardest one anyway.

$\binom{6}{r}(y^2)^r(2y-5)^{6-r}$

$\binom{6}{1}(y^2)(2y-5)^{5}$

So now I need to find the coefficient of  y^4   in   $(2y-5)^{5}$

That will be 

  $\binom{5}{2}(2y)^2(-5)^3\\ =10*4*-125y^2\\ =-5000y^2$

so the y^4 term will be

$\binom{6}{1}(y^2)(-5000y^2)\\ =-30000y^4$

Now you have to do y=0 and y=2   and add them all together.

Stationary point when f'(x)=0

Its a max if  f"(x)<0  and its a min if   f"(x)>0

so you are expected to differentiate and plug in the points.

f(x)=(x²-1)/(x²+3x+2)

$f(x)=\frac{(x²-1)}{(x²+3x+2)}\\ f(x)=\frac{(x-1)(x+1)}{(x+2)(x+1)}\\ \text{there is a hole at x=1 because you cannot divide by 0}\\ f(x)=\frac{(x-1)}{(x+2)}\\ \text{There is an asymptote at x=-2}$

You can finish it :)

Let the thr peices be these lengths      x,  6-y  and y-x

if you add those together you get a total length of 6 units

You know that each of those must be between  0 and 6 units long and that   y>x

Graph the intersection of all those on a nu7mber plane.  The area represents the sample space.

now graph where all the pieces are <5 units   Find the area of that bit.  That is the desired event area

now put the desired event area over the sample space area and you have your answer.

Hi Juriemagic.

You do not have enought points to get an exact answer for your senario.

(-1/3;0), Q(1;0) and R(0;1)

f(x) is a cubic so we need another point, preferably another root

$f'(x)=mx^2+nx+k\\ integrate\\ f(x)=\frac{m}{3}x^3+\frac{n}{2}x^2+kx+t\\ y=\frac{m}{3}x^3+\frac{n}{2}x^2+kx+t\\ $

Sub in the 3 points and you can get all the eunknown constants in terms of just one on them.

OR

alternatively

$y=g(x+\frac{1}{3})(x-1)(x-a)\\ sub \;in\; (0,1) \\ 1=g(+\frac{1}{3})(-1)(-a)\\ 1=g(\frac{a}{3})\\ g=\frac{3}{a}\\ so\\ y=\frac{3}{a}(x+\frac{1}{3})(x-1)(x-a)\\ $

Note :  I have not checked this for careless error. 

Hi Eggrick2011, thanks for that well explained answer.

Unfortunately you made a careless error.  Could you fix it please?

Hi Tiggsy,

Maybe other people have had a go. 

Anyway I enjoyed it   :)

It was enjoyable puzzle, and it was good that I could post an answer without giving anything away.

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If anyone wants hints you should ask for them.    

We always like seeing interest in our posts even if they seem too hard !

Hi Juriemagic  :)

y is the graph of the first derivate, not of the original function so you can ignore all the derivative notation

Graph  $y=f'(x)=mx^2+nx+k$   goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.

becomes

Graph  $y=mx^2+nx+k$   goes through the points P(-1/3;0), Q(1;0) and R(0;1). Detremine the values of m, n and k.

There are at most 2 roots becasue the highest power of x is 2

P and Q are the roots because the y values are 0   so you have 

$y=t(x+\frac{1}{3})(x-1)$      where t is some constant.

Now you use the 3rd point  R(0,1) to find  t

$1=t(0+\frac{1}{3})(0-1)\\ 1=t(\frac{1}{3})(-1)\\ 1=t*-\frac{1}{3}\\ t=-3$

So the function is     $y=-3(x+\frac{1}{3})(x-1)$

Now expand that our to find  m, n  and k

At the end substitute P, Q and R in to check it is correct.   If not go find your, or mine, error.