f4 maths ..急!

Method one:

y = 6 / (4x^2 - 8x+7)
y(4x^2 - 8x + 7) = 6
4yx^2 - 8yx + 7y - 6 = 0
Consider above as quadratic in x, for real value of x,
Discriminant = (8y)^2 - 4(4y)(7y-6) >= 0
64y^2 - 16y(7y-6) >= 0
16y(4y - 7y + 6) >= 0
16y(6 - 3y) >= 0
(y>=0 and 6-3y>= 0) or (y=<0 and 6-3y=<0)
(y>=0 and 2>=y) or (y<=0 and 2<=y)
0 <= y <= 2 the min is 0 and the max is 2
Method two:
y = 6/[4(x^2 - 2x + 1) + 7 - 4]
y = 6/[4(x-1)^2 + 3]
Minimum is when x approaches +/- infinity so min is 0
Maximum is when x = 1 so max = 6/3 = 2

consider 4x^2 - 8x + 7 = 4(x^2 - 2x +7/4) = 4[(x-1)^2 +3/4]
= 4(x-2)^2 +3
therefore the min value of 4x^2 - 8x + 7 is 3 (when x = 2);

Hence the maxmium value of 6/(4x^2 - 8x + 7) is 6/3 = 2.