What is the extreme y=e^x-3e^{-x}-4x ?

The extreme y=e^x-3e^{-x}-4x is Maximum(0,-2),Minimum(ln(3),-4ln(3)+2)

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extreme y=e^x-3e^{-x}-4x

extreme points

y = e^{x} - 3 e^{- x} - 4 x

Maximum (0, - 2), Minimum (ln (3), - 4 ln (3) + 2)

Solution steps

f^{'} (x) = e^{x} + 3 e^{- x} - 4

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < ln (3),

Increasing

: ln (3) < x < \infty

Plug

x = 0

into

e^{x} - 3 e^{- x} - 4 x : - 2

Maximum (0, - 2)

Plug

x = ln (3)

into

e^{x} - 3 e^{- x} - 4 x : - 4 ln (3) + 2

Minimum (ln (3), - 4 ln (3) + 2)

Maximum (0, - 2), Minimum (ln (3), - 4 ln (3) + 2)

f (x) = - 6 x^{3} - 54 x^{2} - 144 x - 90

f (x, y) = 25 - x^{2} - y^{2}

f (x) = (\frac{x + y - x y}{x - y + x y})

3 (x + 2) (x - 10)

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

What is the extreme y=e^x-3e^{-x}-4x ?
The extreme y=e^x-3e^{-x}-4x is Maximum(0,-2),Minimum(ln(3),-4ln(3)+2)