Indicate the regions of the plane

[A_I_]

please help!

i have in two days a maths EXAM, it's mid term exam,
this year I am graduating, I am in last year high school
(maths specialisation, french program)

the problem is that we have a function
f(x)= x2+2x-3/x


M(p,q) p and q are coordinates of M given that p different from 2q.

a) which relation has to verify p and q for which function
y= g(x)= x2+2x-3/px2+qx is monotonous over its interval where it is defined?

b) indicate the regions of the plane wher epoint M has to be put for that the relation above is verified?


please i do need a quick reply
thanks :)
joe

 

[dextercioby]

i have in two days a maths EXAM, it's mid term exam,
this year I am graduating, I am in last year high school
(maths specialisation, french program)

the problem is that we have a function
f(x)= x2+2x-3/x


M(p,q) p and q are coordinates of M given that p different from 2q.

a) which relation has to verify p and q for which function
y= g(x)= x2+2x-3/px2+qx is monotonous over its interval where it is defined?

b) indicate the regions of the plane wher epoint M has to be put for that the relation above is verified?


please i do need a quick reply
thanks :)
joe

Here's what i make of it:$f(x)=x^2+2x-\frac{3}{x}$.
a)$y=g(x)=x^2+(2+q)x-\frac{3}{2px}$.In the case of this weird function,it should be clear that it is defined anywhere,but in the point "x=0",since to me "x" is involved as well in the denominator,so it has to be different from 0.Compute the derivative of "g" wrt to "x" and study its sign in the "g's" domain of definiton.Impose constant sign on the derivative.I believe it's "plus".From there u should be able to pick up some constraints of "p" and "q" besides,the one given that $p\neq 2q$.
b)From solving point a),u have automatically found the geometrical locus of M.

 

[A_I_]

first of all, the function isn't as u understood,
because i don't have the maths program over here.

it is y=f(x)= (x2 +2x -3)/x
and y = g(x)= (x2 +2x -3) / (px2 + qx)

second, i did found the derivative of g(x)

it is: (p -2q)x2 -6p +3q.

i did the delta/
delta prime = 9p2 -3q2 +6pq

when is it negative, when positive??

i figured out later that delta prime is the equation of a hyperbola
but stil i don't know the answer,

if someone can show me a detailed reasoning,
i would appreciate it
btw thanks dextercioby for ur assistance.


PS: if u find any wrong terms, pardon me,
im french educated
im trying my best to translate correctly :)

 

[HallsofIvy]

First, you don't need a "maths program" to put in parentheses. Also, if you are not using Tex, you can represent powers by "^". That is x² can be written x^2.

g' is not (p -2q)x2 -6p +3q. For one thing you are missing the denominator! Of course, that's not important- the denominator is squared so is irrelevant to the question of where g is monotone. However, it looks to me like you have lost a sign:
I get (q- 2p)x²+ 6px+ 3q for the numerator. In order that g be monotone, g' must always have the same sign which essentiall means that it cannot be 0. Setting (q- 2p)x²+ 6px+ 3q = 0 if that is never, true the equation must have only complex roots: in other words the "discrimant" in the quadratic equation, (6p)² - 4(q-2p)(3q)= 36p²- 12q²+ 24pq must be less than 0: 36p²- 12q²+ 24pq < 0. Although it's non-trivial to show it, 36p²- 12q²+ 24pq = 0 (which is basically your 9p² -3q² +6pq= 0) is a hyperbola in the "pq-plane".

 

[A_I_]

so: 36p^2 -12q^2 +24q < 0
is the relation?