What is the extreme f(x)=x^2+((4-x)(4-x)) ?

The extreme f(x)=x^2+((4-x)(4-x)) is Minimum(2,8)

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extreme f(x)=x^2+((4-x)(4-x))

extreme points

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

Minimum (2, 8)

Solution steps

f^{'} (x) = 4 x - 8

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

x^{2} + (4 - x) (4 - x) : 8

Minimum (2, 8)

40000 x + 30000 y - 8 x^{2} - 15 y^{2} - 10 x y

f (x) = \frac{- x ^{2}}{100} + x

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

f (x) = x (15 - 48 + 2 x) (\frac{48}{2} - x)

7 sin (x) + 7 cos (x)

What is the extreme f(x)=x^2+((4-x)(4-x)) ?
The extreme f(x)=x^2+((4-x)(4-x)) is Minimum(2,8)