What is the extreme f(x,y)=e^{2x}-x^3y^5 ?

The extreme f(x,y)=e^{2x}-x^3y^5 is None

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extreme f(x,y)=e^{2x}-x^3y^5

extreme points

f (x, y) = e^{2 x} - x^{3} y^{5}

None

Solution steps

Find the critical points:

None

e^{2 x} - x^{3} y^{5}

has no critical points, therefore there are no extreme points

None

f (x) = 3 x^{4} + x - 1

x + \frac{108}{x ^{2}}

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

f (x) = 5 x - ln (x)

f (x) = x^{4} - 12 x^{3} + 16 x^{2} + 5

What is the extreme f(x,y)=e^{2x}-x^3y^5 ?
The extreme f(x,y)=e^{2x}-x^3y^5 is None