What is the extreme f(t)=(4-t)4^t ?

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

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extreme f(t)=(4-t)4^t

extreme points

f (t) = (4 - t) 4^{t}

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:

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

Domain of

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

Find intervals:

Increasing

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

Decreasing

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

Plug the extreme point

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

into

(4 - t) 4^{t} \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) = 2 x^{3} - 15 x^{2} - 30

f (x) = 15 x^{2} - 15 x^{4}

6 x^{3} - 9 x^{2} - 216 x + 9

f (x) = \frac{x ^{4}}{4} + \frac{5}{3} x^{3} - 5 x^{2} - 50 x + 1

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

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