What is the extreme f(x)=sqrt(x)ln(4x),(0,infinity) ?

The extreme f(x)=sqrt(x)ln(4x),(0,infinity) is Minimum(1/(4e^2),-1/e)

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extreme f(x)=sqrt(x)ln(4x),(0,infinity)

extreme points

f (x) = x ln (4 x), (0, \infty)

Minimum (\frac{1}{4 e ^{2}}, - \frac{1}{e})

Solution steps

Find the critical points:

x = \frac{1}{4 e ^{2}}

Domain of

x ln (4 x), (0, \infty) : x > 0

Find intervals:

Decreasing

: 0 < x < \frac{1}{4 e ^{2}},

Increasing

: \frac{1}{4 e ^{2}} < x < \infty

Plug the extreme point

x = \frac{1}{4 e ^{2}}

into

x ln (4 x) \Rightarrow y = - \frac{1}{e}

Minimum (\frac{1}{4 e ^{2}}, - \frac{1}{e})

- 8 x^{3} + 24 x + 9

f (x) = x^{3} - 3 x^{2} - 45 x + 25

f (x) = x^{3} - 3 x^{2} - 45 x + 22

- 8 x^{3} + 24 x + 3

f (x) = 2 x^{4} - 36 x^{2} - 1, - 4 \leq x \leq 4

What is the extreme f(x)=sqrt(x)ln(4x),(0,infinity) ?
The extreme f(x)=sqrt(x)ln(4x),(0,infinity) is Minimum(1/(4e^2),-1/e)