What is the extreme f(x)=ln(y-x^2)+ln(y-2x) ?

The extreme f(x)=ln(y-x^2)+ln(y-2x) is Maximum(1, 3/2)

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extreme f(x)=ln(y-x^2)+ln(y-2x)

extreme points

f (x) = ln (y - x^{2}) + ln (y - 2 x)

Maximum (1, \frac{3}{2})

Solution steps

Find the critical points:

(1, \frac{3}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{- x ^{4} + 2 x ^{2} y - 4 x ^{2} + 4 x y - 2 y ^{2}}{( y - x ^{2} ) ^{2} ( y - 2 x ) ^{2}}

D (x, y) = \frac{8 x ^{8} - 8 x ^{7} y - 32 x ^{7} + 2 x ^{6} y ^{2} + 24 x ^{6} y - 16 x ^{6} + 112 x ^{5} y - 2 x ^{4} y ^{3} - 116 x ^{4} y ^{2} + 40 x ^{3} y ^{3} - 64 x ^{3} y ^{2} - 4 x ^{2} y ^{4} + 80 x ^{2} y ^{3} + 16 x ^{2} y ^{2} - 32 x y ^{4} - 16 x y ^{3} + 4 y ^{5} + 4 y ^{4}}{( - x ^{2} + y ) ^{4} ( y - 2 x ) ^{4}}

Check the sign of at each critical point

Check critical point

(1, \frac{3}{2}) :

Maximum

Maximum (1, \frac{3}{2})

f (x) = 10000 + 2500 x - 5 x^{2}

f (x) = 2 x^{2} + y^{3} - 6 x y + 2

f (x) = 4 x^{2} + 3

f (x) = (\frac{1}{12} x^{3} - 9 x + 5)

f (x) = 2 x^{2} + 18 x + 16

What is the extreme f(x)=ln(y-x^2)+ln(y-2x) ?
The extreme f(x)=ln(y-x^2)+ln(y-2x) is Maximum(1, 3/2)