Question

Topic: Functions & Graphs.
I could do part (a) but fail to continue on part (b). Any help will be appreciated. Below are problem and attempted answer:
Problem: https://dl.dropbox.com/u/63664351/Mathematics%20at%20A-level/Graphs%20and%20Functions.pdf
Solution part (a): https://dl.dropbox.com/u/63664351/Mathematics%20at%20A-level/Graphs%20and%20Functions%20Solution.pdf

Answer 1

second link is dead

Answer 2

you can attach the pdf here though

Answer 3

Sorry. It is up and good now :))

Answer 4

ok one sec

Answer 5

a) looks perfect

are you completely stumped on b) or do you have something worked out for it?

Answer 6

I am completely lost. Please help

Answer 7

ok one sec

Answer 8

the idea with part b) is that you have to derive f(x) to get f ' (x)

You then use the derivative function to find the extrema. You set the derivative equal to zero and solve for x. From there, you plug each solution into f(x) to find the extrema values.


So

f(x) = (3x^2 + px+3)/(3x^2 + p)

f ' (x) = ((6x + p)*(3x^2 + p) - (6x)*(3x^2+px))/((3x^2+p)^2)

f ' (x) = (3x^2(6x+p) + p(6x+p) - 18x^3 - 6px^2)/((3x^2+p)^2)

f ' (x) = (18x^3+3px^2 + 6px+p^2 - 18x^3 - 6px^2 - 18x)/((3x^2+p)^2)

f ' (x) = (6px+p^2 - 3px^2 - 18x)/((3x^2+p)^2)

f ' (x) = ( -3px^2 + (6p-18)x + p^2 )/((3x^2+p)^2)


Yes that's a lot and it unfortunately gets worse because you now have to find the roots of this function


f ' (x) = (-3px^2 + (6p-18)x + p^2)/((3x^2+p)^2)


0 = (-3px^2 + (6p-18)x + p^2)/((3x^2+p)^2)


-3px^2 + (6p-18)x + p^2 = 0


You then use the quadratic formula to solve for x


x = (-b +- sqrt(b^2 - 4ac) )/(2a)

x = (-(6p-18) +- sqrt((6p-18)^2 - 4(-3p)(p^2)) )/(2(-3p))

x = (-6p+18 +- sqrt(36p^2 - 216p + 324 + 12p^3) )/(-6p)


Once you have those x values, you plug them into f(x) and you add. Then you set that result equal to 4 and solve for p.


There's probably a much easier way to do this, but I'm not seeing it at the moment.

Answer 9

@jim_thompson5910 what is the question?

Answer 10

in here

https://dl.dropbox.com/u/63664351/Mathematics%20at%20A-level/Graphs%20and%20Functions.pdf

part b

Answer 11

Anyone can help with simpler solution?

Answer 12

whoops... check that..., i got p=1, by guess and check...
so the function will be: \(\large f(x)=\frac{3x^2+x+3}{3x^2+1} \)

the two roots obtained as described by @jim_thompson5910 is:
\(\large x_1=\frac{-6+\sqrt{39}}{3} \) and \(\large x_2=\frac{-6-\sqrt{39}}{3} \)

notice that \(\large f(x_1)+f(x_2) = 4 \)

Answer 13

while you're online, hey what do you think @satellite73 ?

Answer 14

i think i am lost because i have no idea what "total sum of extreme value of the function" means in plain english
go with the answer above, looks like they know