What is the extreme f(x)=3xln(x)-8x,x>0 ?

The extreme f(x)=3xln(x)-8x,x>0 is Minimum(e^{5/3},-3e^{5/3})

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extreme f(x)=3xln(x)-8x,x>0

extreme points

f (x) = 3 x ln (x) - 8 x, x > 0

Minimum (e^{\frac{5}{3}}, - 3 e^{\frac{5}{3}})

Solution steps

Find the critical points:

x = e^{\frac{5}{3}}

Domain of

3 x ln (x) - 8 x, x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < e^{\frac{5}{3}},

Increasing

: e^{\frac{5}{3}} < x < \infty

Plug the extreme point

x = e^{\frac{5}{3}}

into

3 x ln (x) - 8 x \Rightarrow y = - 3 e^{\frac{5}{3}}

Minimum (e^{\frac{5}{3}}, - 3 e^{\frac{5}{3}})

\frac{2 x}{5 x + 8}

- 2 x^{2} + 2

f (x) = 3 x^{3} + 7 x^{2} + 9 x + 12

f (x) = \frac{( 5 + x ^{4} )}{( x ^{2} - x ^{4} )}

(x + 2)^{2}

What is the extreme f(x)=3xln(x)-8x,x>0 ?
The extreme f(x)=3xln(x)-8x,x>0 is Minimum(e^{5/3},-3e^{5/3})