What is the extreme f(x)=x^2ln(x), 1/3 <= x<= 5/6 ?

The extreme f(x)=x^2ln(x), 1/3 <= x<= 5/6 is

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extreme f(x)=x^2ln(x), 1/3 <= x<= 5/6

extreme points

f (x) = x^{2} ln (x), \frac{1}{3} \leq x \leq \frac{5}{6}

Maximum (\frac{1}{3}, - \frac{1}{9} ln (3)), Minimum (\frac{1}{e}, - \frac{1}{2 e}), Maximum (\frac{5}{6}, \frac{25}{36} ln (\frac{5}{6}))

Solution steps

Find the critical points:

x = \frac{1}{3}, x = \frac{1}{e}, x = \frac{5}{6}

Domain of

x^{2} ln (x), \frac{1}{3} \leq x \leq \frac{5}{6} : \frac{1}{3} \leq x \leq \frac{5}{6}

Find intervals:

Decreasing

: \frac{1}{3} < x < \frac{1}{e},

Increasing

: \frac{1}{e} < x < \frac{5}{6}

Plug the extreme point

x = \frac{1}{3}

into

x^{2} ln (x) \Rightarrow y = - \frac{1}{9} ln (3)

Maximum (\frac{1}{3}, - \frac{1}{9} ln (3))

Plug the extreme point

x = \frac{1}{e}

into

x^{2} ln (x) \Rightarrow y = - \frac{1}{2 e}

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

Plug the extreme point

x = \frac{5}{6}

into

x^{2} ln (x) \Rightarrow y = \frac{25}{36} ln (\frac{5}{6})

Maximum (\frac{5}{6}, \frac{25}{36} ln (\frac{5}{6}))

Maximum (\frac{1}{3}, - \frac{1}{9} ln (3)), Minimum (\frac{1}{e}, - \frac{1}{2 e}), Maximum (\frac{5}{6}, \frac{25}{36} ln (\frac{5}{6}))

f (x) = - x^{4} + 6 x^{2}

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

f (x) = t 9 - t^{2}

f (x) = 2 x^{3} - 6 x^{2} - 2 x + 5

f (x) = x^{2} - 5 x

What is the extreme f(x)=x^2ln(x), 1/3 <= x<= 5/6 ?
The extreme f(x)=x^2ln(x), 1/3 <= x<= 5/6 is