What is the extreme f(x)=-0.020724x^2+0.3692x-0.6287 ?

The extreme f(x)=-0.020724x^2+0.3692x-0.6287 is Maximum((0.3692)/(0.041448),1.01563…)

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extreme f(x)=-0.020724x^2+0.3692x-0.6287

extreme points

f (x) = - 0.020724 x^{2} + 0.3692 x - 0.6287

Maximum (\frac{0.3692}{0.041448}, 1.01563 \dots)

Solution steps

f^{'} (x) = - 0.041448 x + 0.3692

Find intervals:

Increasing

: - \infty < x < \frac{0.3692}{0.041448},

Decreasing

: \frac{0.3692}{0.041448} < x < \infty

Plug

x = \frac{0.3692}{0.041448}

into

- 0.020724 x^{2} + 0.3692 x - 0.6287 : 1.01563 \dots

Maximum (\frac{0.3692}{0.041448}, 1.01563 \dots)

f (x, y) = x^{3} - y^{2} - 3 x + 12 y

f (x) = (x^{2} - 12) \cdot e^{- 2 x}

f (x) = 49 x^{2} - 10 x + 17

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

f (x, y) = - 3 y^{3} - 3 x y + 13 y - 2 \cdot 2 + 5 x + 19

What is the extreme f(x)=-0.020724x^2+0.3692x-0.6287 ?
The extreme f(x)=-0.020724x^2+0.3692x-0.6287 is Maximum((0.3692)/(0.041448),1.01563…)