What is the extreme f(x)=sqrt(x^2+0.3x)-0.22x ?

The extreme f(x)=sqrt(x^2+0.3x)-0.22x is Minimum(-3/10 ,0.066),Minimum(0,0)

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extreme f(x)=sqrt(x^2+0.3x)-0.22x

extreme points

f (x) = x^{2} + 0.3 x - 0.22 x

Minimum (- \frac{3}{10}, 0.066), Minimum (0, 0)

Solution steps

Find the critical points:

x = - \frac{3}{10}, x = 0

Domain of

x^{2} + 0.3 x - 0.22 x : x \leq - 0.3 or x \geq 0

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{10},

Constant

: E m pt y I n t er v a l,

Increasing

: 0 < x < \infty

Plug the extreme point

x = - \frac{3}{10}

into

x^{2} + 0.3 x - 0.22 x \Rightarrow y = 0.066

Minimum (- \frac{3}{10}, 0.066)

Plug the extreme point

x = 0

into

x^{2} + 0.3 x - 0.22 x \Rightarrow y = 0

Minimum (0, 0)

Minimum (- \frac{3}{10}, 0.066), Minimum (0, 0)

What is the extreme f(x)=sqrt(x^2+0.3x)-0.22x ?
The extreme f(x)=sqrt(x^2+0.3x)-0.22x is Minimum(-3/10 ,0.066),Minimum(0,0)