SuicideFuel Problem

[trying to ascend]

[UWSL]Suppose a and b are real and different roots of the equation [/UWSL][UWSL]4x² - 4tx − 1=0 (t ϵ R). The interval [a,b] is the domain of the function[/UWSL]


f(x) 2x - t/x² + 1. Be g(t) = max f(x) - min f(x).

Therefore, the value of g(0) is?

 

[NoBitches]

g(0)=maxf(x)-minf(x)
where f(x)=2x+1 (since t=0)
maxf(x)=2b+1
minf(x)=2a+1
g(0)=2(b-a)
4x^2-1=(2x+1)(2x-1)
So A=-1/2 and B=1/2
g(0)=2

 

[I.N.C.E.L.S. Boss]

G(0) = max F(x) - min F(x)

To determine the max min, derivation :
F(x)/dx = 2 - (-2/x^3(-3)) t = 2 + 2t/x^(-3)
For t=0, F(x)/dx = 2 > 0
F(x) is always ascending for t=0. So, max F(x) = 2b+1 and min F(x) = 2a+1

G(0) = 2b+1 - 2a -1
= 2(b-a)

let's get this b-a = ?
a and b are solution to the equation 4x²-4tx-1 = 0
when t=0, 4x²-1 = 0

So, 4a² = 1 and 4b² = 1
therefore, 4b²-4a² = 0
4(b²-a²) = 0
b²-a² = 0
(b-a)(b+a) = 0
So either b-a = 0 or b+a = 0
so either b=a or b=-a

You said "Suppose a and b are different roots of the equation", so a and b cannot be equal.
The only solution is b = -a

G(0) = 2(-a-a) = -4a

 

[I.N.C.E.L.S. Boss]

G(0) = max F(x) - min F(x)

To determine the max min, derivation :
F(x)/dx = 2 - (-2/x^3(-3)) t = 2 + 2t/x^(-3)
For t=0, F(x)/dx = 2 > 0
F(x) is always ascending for t=0. So, max F(x) = 2b+1 and min F(x) = 2a+1

G(0) = 2b+1 - 2a -1
= 2(b-a)

let's get this b-a = ?
a and b are solution to the equation 4x²-4tx-1 = 0
when t=0, 4x²-1 = 0

So, 4a² = 1 and 4b² = 1
therefore, 4b²-4a² = 0
4(b²-a²) = 0
b²-a² = 0
(b-a)(b+a) = 0
So either b-a = 0 or b+a = 0
so either b=a or b=-a

You said "Suppose a and b are different roots of the equation", so a and b cannot be equal.
The only solution is b = -a

G(0) = 2(-a-a) = -4a

a and -a are, therefore, both solutions to the equation 4x²-1 =0
4(-a)²-1 = 4a²-1 = 0
a² = 1/4
a = 1/2 or a= -1/2
thus, b = -1/2 or b = 1/2

Since b>a (the interval [a,b]), the only possible solution is b = 1/2 and a = -1/2

G(0) = -4a = -4 * -1/2 = 2
G(0) = 2


NOW GIVE ME MY GOOD BOY POINTS