What is the extreme f(x)=c(x)=x^2-4000x+50 ?

The extreme f(x)=c(x)=x^2-4000x+50 is Minimum(2000,-3999950)

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extreme f(x)=c(x)=x^2-4000x+50

extreme points

f (x) = c (x) = x^{2} - 4000 x + 50

Minimum (2000, - 3999950)

Solution steps

f^{'} (x) = 2 x - 4000

Find intervals:

Decreasing

: - \infty < x < 2000,

Increasing

: 2000 < x < \infty

Plug

x = 2000

into

x^{2} - 4000 x + 50 : - 3999950

Minimum (2000, - 3999950)

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

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

f (x) = - x^{3} + 3 x - 5

f (x) = \frac{2 x + 5}{3}, (0, 5)

f (x) = \frac{( 1 )}{( 3 ) x ^{(3)} - 400 x}

What is the extreme f(x)=c(x)=x^2-4000x+50 ?
The extreme f(x)=c(x)=x^2-4000x+50 is Minimum(2000,-3999950)