What is the extreme f(x)=2600+2x+0.001x^2 ?

The extreme f(x)=2600+2x+0.001x^2 is Minimum(-1000,1600)

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extreme f(x)=2600+2x+0.001x^2

extreme points

f (x) = 2600 + 2 x + 0.001 x^{2}

Minimum (- 1000, 1600)

Solution steps

f^{'} (x) = 0.002 x + 2

Find intervals:

Decreasing

: - \infty < x < - 1000,

Increasing

: - 1000 < x < \infty

Plug

x = - 1000

into

2600 + 2 x + 0.001 x^{2} : 1600

Minimum (- 1000, 1600)

f (x) = x^{3} - 12 x^{2} - 27 x + 9 (- 2)

f (x) = 13 + 7 x + x^{2}

x^{\frac{1}{3}} (x + 4), - 27 \leq x \leq 27

f (x) = 1 + 4 x - 5 y

f (x) = - 2 x + 5 ln (2 x), (1, 5)

What is the extreme f(x)=2600+2x+0.001x^2 ?
The extreme f(x)=2600+2x+0.001x^2 is Minimum(-1000,1600)