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Popular >

extreme y=(e^{(2-x)}+x^2-x)/2

Solution

extreme points

y = \frac{e ^{(2 - x)} + x ^{2} - x}{2}

Solution

Minimum W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2}, \frac{4 W _{0} ( \frac{e ^{\frac{3}{2}}}{2} ) ^{2} + 4 e ^{\frac{3}{2} - W_{0} (\frac{e ^{\frac{3}{2}}}{2})} - 1}{8}

Solution steps

f^{'} (x) = - \frac{1}{2} e^{2 - x} + x - \frac{1}{2}

Find intervals:

Decreasing

: - \infty < x < W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2},

Increasing

: W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2} < x < \infty

Plug

x = W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2}

into

\frac{e ^{2 - x} + x ^{2} - x}{2} : \frac{4 W _{0} ( \frac{e ^{\frac{3}{2}}}{2} ) ^{2} + 4 e ^{\frac{3}{2} - W_{0} (\frac{e ^{\frac{3}{2}}}{2})} - 1}{8}

Minimum W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2}, \frac{4 W _{0} ( \frac{e ^{\frac{3}{2}}}{2} ) ^{2} + 4 e ^{\frac{3}{2} - W_{0} (\frac{e ^{\frac{3}{2}}}{2})} - 1}{8}

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Frequently Asked Questions (FAQ)

What is the extreme y=(e^{(2-x)}+x^2-x)/2 ?
The extreme y=(e^{(2-x)}+x^2-x)/2 is