What is the extreme f(x)=(3-x)3^x ?

The extreme f(x)=(3-x)3^x is Maximum(-1/(ln(3))+3,(3^{\frac{-1+3ln(3))/(ln(3))}}{ln(3)})

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extreme f(x)=(3-x)3^x

extreme

Maximum (- \frac{1}{ln ( 3 )} + 3, \frac{3 ^{\frac{- 1 + 3 l n ( 3 )}{l n ( 3 )}}}{ln ( 3 )})

Solution steps

f^{'} (x) = 3^{x} (- 1 - ln (3) (x - 3))

Find intervals:

Increasing

: - \infty < x < - \frac{1}{ln ( 3 )} + 3,

Decreasing

: - \frac{1}{ln ( 3 )} + 3 < x < \infty

Plug

x = - \frac{1}{ln ( 3 )} + 3

into

(3 - x) 3^{x} : \frac{3 ^{\frac{- 1 + 3 l n ( 3 )}{l n ( 3 )}}}{ln ( 3 )}

Maximum (- \frac{1}{ln ( 3 )} + 3, \frac{3 ^{\frac{- 1 + 3 l n ( 3 )}{l n ( 3 )}}}{ln ( 3 )})

What is the extreme f(x)=(3-x)3^x ?
The extreme f(x)=(3-x)3^x is Maximum(-1/(ln(3))+3,(3^{\frac{-1+3ln(3))/(ln(3))}}{ln(3)})