What is the extreme f(x)=7ln(3xe^{-x}) ?

The extreme f(x)=7ln(3xe^{-x}) is Maximum(1,7(ln(3)-1))

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extreme f(x)=7ln(3xe^{-x})

extreme points

f (x) = 7 ln (3 x e^{- x})

Maximum (1, 7 (ln (3) - 1))

Solution steps

f^{'} (x) = 7 (\frac{1}{x} - 1)

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

7 ln (3 x e^{- x}) : 7 (ln (3) - 1)

Maximum (1, 7 (ln (3) - 1))

f (x) = 3 x, - 1 \leq x \leq 1

f (x) = \frac{1}{5} x^{3} - x^{2} + 1

\frac{4 x}{x ^{2} - 1}

9 - x^{2}

f (x) = 3 x^{3} + 3 x^{2}

What is the extreme f(x)=7ln(3xe^{-x}) ?
The extreme f(x)=7ln(3xe^{-x}) is Maximum(1,7(ln(3)-1))