What is the extreme f(x)=32e^{4x}x-28e^{4x} ?

The extreme f(x)=32e^{4x}x-28e^{4x} is Minimum(5/8 ,-8e^{5/2})

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extreme f(x)=32e^{4x}x-28e^{4x}

extreme

Minimum (\frac{5}{8}, - 8 e^{\frac{5}{2}})

Solution steps

f^{'} (x) = 16 e^{4 x} (8 x - 5)

Find intervals:

Decreasing

: - \infty < x < \frac{5}{8},

Increasing

: \frac{5}{8} < x < \infty

Plug

x = \frac{5}{8}

into

32 e^{4 x} x - 28 e^{4 x} : - 8 e^{\frac{5}{2}}

Minimum (\frac{5}{8}, - 8 e^{\frac{5}{2}})

What is the extreme f(x)=32e^{4x}x-28e^{4x} ?
The extreme f(x)=32e^{4x}x-28e^{4x} is Minimum(5/8 ,-8e^{5/2})