What is the extreme f(y)=x^{y+1} ?

The extreme f(y)=x^{y+1} is Saddle(-1,1)

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extreme f(y)=x^{y+1}

extreme points

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

Saddle (- 1, 1)

Solution steps

Find the critical points:

(- 1, 1)

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

D (y, x) = - y x^{2 y} ln^{2} (x) - 2 y x^{2 y} ln (x) - x^{2 y} ln^{2} (x) - 2 x^{2 y} ln (x) - x^{2 y}

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Saddle

Saddle (- 1, 1)

f (x) = x^{2} lo g_{8} (x)

x + \frac{192}{x}

f (x) = - \frac{x ^{2}}{10} + 2 x + 3

f (x) = (x^{3} - 3 x^{2} + 3 x + 1)

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