What is the extreme f(x)=2(y+1)^2+y^2-2y-2 ?

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

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extreme f(x)=2(y+1)^2+y^2-2y-2

extreme points

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

Minimum (- \frac{1}{3}, - \frac{1}{3})

Solution steps

f^{'} (y) = 6 y + 2

Find intervals:

Decreasing

: - \infty < y < - \frac{1}{3},

Increasing

: - \frac{1}{3} < y < \infty

Plug

y = - \frac{1}{3}

into

2 (y + 1)^{2} + y^{2} - 2 y - 2 : - \frac{1}{3}

Minimum (- \frac{1}{3}, - \frac{1}{3})

f (x) = x e^{\frac{x}{2}}, - 3 \leq x \leq 1

x + y - x y

p (x) = 3 x^{4} + 2 x^{3} + 5 x^{2} + a x - 2

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

f (x) = \frac{x ^{2}}{9} + 11 + \frac{6}{x}

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