What is the extreme f(x)=-2+e^{3x}(4-2x) ?

The extreme f(x)=-2+e^{3x}(4-2x) is Maximum(5/3 ,-2+(2e^5)/3)

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extreme f(x)=-2+e^{3x}(4-2x)

extreme points

f (x) = - 2 + e^{3 x} (4 - 2 x)

Maximum (\frac{5}{3}, - 2 + \frac{2 e ^{5}}{3})

Solution steps

f^{'} (x) = - 2 e^{3 x} (3 x - 5)

Find intervals:

Increasing

: - \infty < x < \frac{5}{3},

Decreasing

: \frac{5}{3} < x < \infty

Plug

x = \frac{5}{3}

into

- 2 + e^{3 x} (4 - 2 x) : - 2 + \frac{2 e ^{5}}{3}

Maximum (\frac{5}{3}, - 2 + \frac{2 e ^{5}}{3})

\frac{16}{x ^{2} - 2 x - 8}

(- 8.6) (0.1)

f (x) = \frac{x + 6}{x ^{2} + 2}

f (x) = - 7 x - 2

4 x + 8 y = 4

What is the extreme f(x)=-2+e^{3x}(4-2x) ?
The extreme f(x)=-2+e^{3x}(4-2x) is Maximum(5/3 ,-2+(2e^5)/3)