What is the extreme f(x)=(8x-9)e^{4x} ?

The extreme f(x)=(8x-9)e^{4x} is Minimum(7/8 ,-2e^{7/2})

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extreme f(x)=(8x-9)e^{4x}

extreme points

f (x) = (8 x - 9) e^{4 x}

Minimum (\frac{7}{8}, - 2 e^{\frac{7}{2}})

Solution steps

f^{'} (x) = 4 e^{4 x} (8 x - 7)

Find intervals:

Decreasing

: - \infty < x < \frac{7}{8},

Increasing

: \frac{7}{8} < x < \infty

Plug

x = \frac{7}{8}

into

(8 x - 9) e^{4 x} : - 2 e^{\frac{7}{2}}

Minimum (\frac{7}{8}, - 2 e^{\frac{7}{2}})

y = 6 x e^{- x^{2}}

2 x^{2} + y^{2}

f (x) = 3 csc (x)

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

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12

What is the extreme f(x)=(8x-9)e^{4x} ?
The extreme f(x)=(8x-9)e^{4x} is Minimum(7/8 ,-2e^{7/2})