What is the extreme f(x)=0.001x^2+2.4x-90 ?

The extreme f(x)=0.001x^2+2.4x-90 is Minimum(-1200,-1530)

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extreme f(x)=0.001x^2+2.4x-90

extreme points

f (x) = 0.001 x^{2} + 2.4 x - 90

Minimum (- 1200, - 1530)

Solution steps

f^{'} (x) = 0.002 x + 2.4

Find intervals:

Decreasing

: - \infty < x < - 1200,

Increasing

: - 1200 < x < \infty

Plug

x = - 1200

into

0.001 x^{2} + 2.4 x - 90 : - 1530

Minimum (- 1200, - 1530)

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

g (x) = 6 x^{4} - 32 x^{3} + 48 x^{2} - 7

f (x) = tan (\frac{π x}{20}), 0 \leq x \leq 5

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

f (x) = - x^{2} + 128 ln (x), 1 \leq x \leq 12

What is the extreme f(x)=0.001x^2+2.4x-90 ?
The extreme f(x)=0.001x^2+2.4x-90 is Minimum(-1200,-1530)