What is the extreme f(x)=0.4x^2+110.2x,0<= x ?

The extreme f(x)=0.4x^2+110.2x,0<= x is Minimum(0,0)

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extreme f(x)=0.4x^2+110.2x,0<= x

extreme points

f (x) = 0.4 x^{2} + 110.2 x, 0 \leq x

Minimum (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

0.4 x^{2} + 110.2 x, 0 \leq x : x \geq 0

Find intervals:

Increasing

: 0 < x < \infty

Plug the extreme point

x = 0

into

0.4 x^{2} + 110.2 x \Rightarrow y = 0

Minimum (0, 0)

f (x) = \frac{( 3 x )}{( 9 - x ^{2} )}, - 3 \leq x \leq 2

f (x) = \frac{1}{4} x + \frac{1}{x}, x > 0

f (0) = x^{2} - 4 x - 4

\frac{7 e ^{x} + 7 e ^{- x}}{2}

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

What is the extreme f(x)=0.4x^2+110.2x,0<= x ?
The extreme f(x)=0.4x^2+110.2x,0<= x is Minimum(0,0)