What is the extreme f(y_{=3x^2-8x+5})=y_{=3x^2-8x+5}(x) ?

The extreme f(y_{=3x^2-8x+5})=y_{=3x^2-8x+5}(x) is Saddle(0,0)

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extreme f(y_{=3x^2-8x+5})=y_{=3x^2-8x+5}(x)

extreme points

f (y_{= 3 x^{2} - 8 x + 5}) = y_{= 3 x^{2} - 8 x + 5} (x)

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial y _{= 3 x^{2} - 8 x + 5}^{2}} = 0

D (y_{= 3 x^{2} - 8 x + 5}, x) = - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = 4 - 3^{x + 2}

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} - 1

f (x) = 3 cos (x) + 1

y = - \frac{5}{2} x - 1

f (x) = 3 x^{2} + 8 x - 1

What is the extreme f(y_{=3x^2-8x+5})=y_{=3x^2-8x+5}(x) ?
The extreme f(y_{=3x^2-8x+5})=y_{=3x^2-8x+5}(x) is Saddle(0,0)