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 points

f (x) = e^{\frac{x}{2}} (x + y^{2})

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)

f (x) = x^{4} - 50 x^{2} + 1

f (x) = x^{4} - 50 x^{2} + 2

y = x^{3} - 3 x^{2} + 5

+ \frac{x}{x ^{2} - 25}

f (x, y) = x y - 4 x - 4 y - x^{2} - y^{2}

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)