What is the extreme f(x)=e^{-x}+4x ?

The extreme f(x)=e^{-x}+4x is Minimum(-2ln(2),4-8ln(2))

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extreme f(x)=e^{-x}+4x

extreme points

f (x) = e^{- x} + 4 x

Minimum (- 2 ln (2), 4 - 8 ln (2))

Solution steps

f^{'} (x) = - e^{- x} + 4

Find intervals:

Decreasing

: - \infty < x < - 2 ln (2),

Increasing

: - 2 ln (2) < x < \infty

Plug

x = - 2 ln (2)

into

e^{- x} + 4 x : 4 - 8 ln (2)

Minimum (- 2 ln (2), 4 - 8 ln (2))

2 x^{2} - 2 y^{2} - x^{4} - 2 x^{2} y^{2} - y^{4}

f (x) = x - ln (x), 0 \leq x \leq 2

f (x) = x^{3} - 24 ln (x) + 5

f (x) = 16 x + 19 x^{\frac{16}{19}}

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

What is the extreme f(x)=e^{-x}+4x ?
The extreme f(x)=e^{-x}+4x is Minimum(-2ln(2),4-8ln(2))