What is the extreme f(x)=0.0001x^2+3.2x-90 ?

The extreme f(x)=0.0001x^2+3.2x-90 is Minimum(-(3.2)/(0.0002),-25690)

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extreme f(x)=0.0001x^2+3.2x-90

extreme points

f (x) = 0.0001 x^{2} + 3.2 x - 90

Minimum (- \frac{3.2}{0.0002}, - 25690)

Solution steps

f^{'} (x) = 0.0002 x + 3.2

Find intervals:

Decreasing

: - \infty < x < - \frac{3.2}{0.0002},

Increasing

: - \frac{3.2}{0.0002} < x < \infty

Plug

x = - \frac{3.2}{0.0002}

into

0.0001 x^{2} + 3.2 x - 90 : - 25690

Minimum (- \frac{3.2}{0.0002}, - 25690)

f (x, y) = θ

\frac{x ^{3}}{3} - x^{2} - 35 x + 4

f (x, y) = e^{6 x^{2} + 4 y^{2} + 10}

f (x, y) = x y - 8 x - y^{2} + 12 x + 160

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

What is the extreme f(x)=0.0001x^2+3.2x-90 ?
The extreme f(x)=0.0001x^2+3.2x-90 is Minimum(-(3.2)/(0.0002),-25690)