What is the extreme f(x)=0.25x^2+x+2,(x>= 0) ?

The extreme f(x)=0.25x^2+x+2,(x>= 0) is Minimum(0,2)

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extreme f(x)=0.25x^2+x+2,(x>= 0)

extreme points

f (x) = 0.25 x^{2} + x + 2, (x \geq 0)

Minimum (0, 2)

Solution steps

Find the critical points:

x = 0

Domain of

0.25 x^{2} + x + 2, x \geq 0 : x \geq 0

Find intervals:

Increasing

: 0 < x < \infty

Plug the extreme point

x = 0

into

0.25 x^{2} + x + 2 \Rightarrow y = 2

Minimum (0, 2)

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

f (x) = x^{4} - 50 x^{2} + 1, - 4 \leq x \leq 11

f (x) = \frac{16}{x ^{2} + 4}

f (x) = x + sin (x) + cos (x)

\frac{x ^{2}}{( x - 5 ) ^{2}}

What is the extreme f(x)=0.25x^2+x+2,(x>= 0) ?
The extreme f(x)=0.25x^2+x+2,(x>= 0) is Minimum(0,2)