What is the extreme f(x)=0.001x^2+3.2x-10 ?

The extreme f(x)=0.001x^2+3.2x-10 is Minimum(-1600,-2570)

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extreme f(x)=0.001x^2+3.2x-10

extreme points

f (x) = 0.001 x^{2} + 3.2 x - 10

Minimum (- 1600, - 2570)

Solution steps

f^{'} (x) = 0.002 x + 3.2

Find intervals:

Decreasing

: - \infty < x < - 1600,

Increasing

: - 1600 < x < \infty

Plug

x = - 1600

into

0.001 x^{2} + 3.2 x - 10 : - 2570

Minimum (- 1600, - 2570)

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

f (x) = - 7 x^{2} - 7 x y - 6 y^{2} - 35 x - 26 y + 9

3 (x - 4)^{\frac{2}{3}} + 6

y = x^{3} + 3 x^{2}

f (x) = 4 x^{\frac{3}{5}}

What is the extreme f(x)=0.001x^2+3.2x-10 ?
The extreme f(x)=0.001x^2+3.2x-10 is Minimum(-1600,-2570)