What is the extreme f(x)=2xθ^x+θ^x ?

The extreme f(x)=2xθ^x+θ^x is Saddle(0, 1/(e^2))

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extreme f(x)=2xθ^x+θ^x

extreme points

f (x) = 2 x θ^{x} + θ^{x}

Saddle (0, \frac{1}{e ^{2}})

Solution steps

Find the critical points:

(0, \frac{1}{e ^{2}})

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

D (x, θ) = x θ^{2 x - 2} ln (θ) (2 x ln (θ) + ln (θ) + 4) (- 1 + x) (2 x + 1) - θ^{2 (x - 1)} (2 x^{2} ln (θ) + x ln (θ) + 4 x + 1)^{2}

Check the sign of at each critical point

Check critical point

(0, \frac{1}{e ^{2}}) :

Saddle

Saddle (0, \frac{1}{e ^{2}})

27.8

P (x, y) = 40 x + 50 y

y = \frac{x ^{2} + 1}{x}

f (x) = x^{3} - 12 x^{2} - 27 x + 2, - 2 \leq x \leq 10

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

What is the extreme f(x)=2xθ^x+θ^x ?
The extreme f(x)=2xθ^x+θ^x is Saddle(0, 1/(e^2))