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) \cdot 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) \cdot 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) \cdot 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 )})

f (x) = 3 + 3 x - 3 x^{2}, 0 \leq x \leq 3

100

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

x^{2} + \frac{405000}{x}

\frac{2 x ^{2} - 14 x + 24}{x ^{2} + 6 x + 40}

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)})