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…)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

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)

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…)