What is the extreme f(x)=e^{x/2}(x+y^2) ?

The extreme f(x)=e^{x/2}(x+y^2) is Minimum(-2,0)

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extreme f(x)=e^{x/2}(x+y^2)

extreme

Minimum (- 2, 0)

Solution steps

Find the critical points:

(- 2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{\frac{x}{2}}

D (x, y) = \frac{- e ^{x} y ^{2} + 4 e ^{x} + e ^{x} x}{2}

Check the sign of at each critical point

Check critical point

(- 2, 0) :

Minimum

Minimum (- 2, 0)

What is the extreme f(x)=e^{x/2}(x+y^2) ?
The extreme f(x)=e^{x/2}(x+y^2) is Minimum(-2,0)