What is the extreme 3(e^{-0.8t}-e^{-1.2t}) ?

The extreme 3(e^{-0.8t}-e^{-1.2t}) is Minimum((ln(1.2)-ln(0.8))/(0.4),0.44444…)

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extreme 3(e^{-0.8t}-e^{-1.2t})

extreme points

3 (e^{- 0.8 t} - e^{- 1.2 t})

Minimum (\frac{ln ( 1.2 ) - ln ( 0.8 )}{0.4}, 0.44444 \dots)

Solution steps

f^{'} (t) = 3 (- 0.8 e^{- 0.8 t} + 1.2 e^{- 1.2 t})

Find intervals:

Increasing

: - \infty < t < ln (\frac{3}{2}),

Decreasing

: ln (\frac{3}{2}) < t < \infty

Plug

t = \frac{ln ( 1.2 ) - ln ( 0.8 )}{0.4}

into

3 (e^{- 0.8 t} - e^{- 1.2 t}) : 0.44444 \dots

Minimum (\frac{ln ( 1.2 ) - ln ( 0.8 )}{0.4}, 0.44444 \dots)

f (x) = 6 x^{2} - 8 x + 2

f (x) = ∣ 2 x + 1 ∣

\frac{1}{∣ x ∣ + 1}

f (x, y) = e^{2 x} (x + y^{2} + 2 y) + 10

P (x, y) = x^{2} - y^{2}

What is the extreme 3(e^{-0.8t}-e^{-1.2t}) ?
The extreme 3(e^{-0.8t}-e^{-1.2t}) is Minimum((ln(1.2)-ln(0.8))/(0.4),0.44444…)