What is the extreme f(x)=4(z^{-x})+1 ?

The extreme f(x)=4(z^{-x})+1 is Saddle(0,1)

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

extreme points

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

Saddle (0, 1)

Solution steps

Find the critical points:

(0, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4 x z^{- x - 2} (- x - 1)

D (x, z) = - 16 x z^{- 2 x - 2} ln^{2} (z) (- x - 1) - 16 (- x z^{- x - 1} ln (z) + z^{- x - 1})^{2}

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Saddle (0, 1)

y = 15 x e^{- x}

f (x) = x^{2} + 6

f (x) = 6 x^{2} + 1 - x, 0 \leq x \leq 3

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

5 x^{3} + 67.5 x^{2} + 2 x + 10

What is the extreme f(x)=4(z^{-x})+1 ?
The extreme f(x)=4(z^{-x})+1 is Saddle(0,1)