What is the extreme y=(4-x)4^x ?

The extreme y=(4-x)4^x is Maximum(-1/(2ln(2))+4,(4^{\frac{-1+8ln(2))/(2ln(2))}}{2ln(2)})

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extreme y=(4-x)4^x

extreme points

y = (4 - x) 4^{x}

Maximum (- \frac{1}{2 ln ( 2 )} + 4, \frac{4 ^{\frac{- 1 + 8 l n ( 2 )}{2 l n ( 2 )}}}{2 ln ( 2 )})

Solution steps

Find the critical points:

x = - \frac{1}{2 ln ( 2 )} + 4

Domain of

(4 - x) 4^{x} : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - \frac{1}{2 ln ( 2 )} + 4,

Decreasing

: - \frac{1}{2 ln ( 2 )} + 4 < x < \infty

Plug the extreme point

x = - \frac{1}{2 ln ( 2 )} + 4

into

(4 - x) 4^{x} \Rightarrow y = \frac{4 ^{\frac{- 1 + 8 l n ( 2 )}{2 l n ( 2 )}}}{2 ln ( 2 )}

Maximum (- \frac{1}{2 ln ( 2 )} + 4, \frac{4 ^{\frac{- 1 + 8 l n ( 2 )}{2 l n ( 2 )}}}{2 ln ( 2 )})

- 8 t^{2} + 36 t + 99

f (x) = 2 x^{4} - 2 x^{3}

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

f (x) = 6 x^{3} - 9 x^{2} - 216 x + 3

f (x, y) = 338 y^{2} + x^{2} - x^{2} y

What is the extreme y=(4-x)4^x ?
The extreme y=(4-x)4^x is Maximum(-1/(2ln(2))+4,(4^{\frac{-1+8ln(2))/(2ln(2))}}{2ln(2)})