What is the extreme f(x)=3e^{x/2}-(x+1)sqrt(x+1) ?

The extreme f(x)=3e^{x/2}-(x+1)sqrt(x+1) is Saddle(0,2)

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extreme f(x)=3e^{x/2}-(x+1)sqrt(x+1)

extreme points

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

Saddle (0, 2)

Solution steps

f^{'} (x) = \frac{3 e ^{\frac{x}{2}} - 3 ( x + 1 ) ^{\frac{1}{2}}}{2}

Find intervals:

Increasing

: - 1 \leq x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

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

Saddle (0, 2)

f (x) = \frac{e ^{x}}{9 + e ^{x}}

6 x y - 1.2 x^{2} - 3 y^{4}

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

\frac{2 x + 1}{x - 3}

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

What is the extreme f(x)=3e^{x/2}-(x+1)sqrt(x+1) ?
The extreme f(x)=3e^{x/2}-(x+1)sqrt(x+1) is Saddle(0,2)