What is the extreme y=((Ix+2I-1)/(2x+3))e^{2x} ?

The extreme y=((Ix+2I-1)/(2x+3))e^{2x} is Saddle(4,-2)

extreme y=((Ix+2I-1)/(2x+3))e^{2x}

extreme points

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

Saddle (4, - 2)

Solution steps

Find the critical points:

(4, - 2)

\frac{\partial ^{2} f}{\partial I ^{2}} = \frac{4 e ^{2 x} ( 4 I x ^{3} + 20 I x ^{2} - 4 x ^{2} + 31 I x - 8 x + 16 I - 5 )}{( 2 x + 3 ) ^{3}}

D (I, x) = - \frac{( 4 e ^{2 x} x ^{2} + 14 e ^{2 x} x + 11 e ^{2 x} ) ^{2}}{( 2 x + 3 ) ^{4}}

Check the sign of at each critical point

Check critical point

(4, - 2) :

Saddle

Saddle (4, - 2)

x + \frac{3}{2} x^{\frac{2}{3}}

+ ∣ x^{2} - 3 x + 2 ∣

f (x) = (- 5 x^{2} + 10 x y - 20 x - 7 y^{2} + 240 y - 5300)

f (x, y) = 3 x y + 6 y - 5 x

y = I n ∣ 1 - x ∣

What is the extreme y=((Ix+2I-1)/(2x+3))e^{2x} ?
The extreme y=((Ix+2I-1)/(2x+3))e^{2x} is Saddle(4,-2)