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LenaWriter [7]
2 years ago
11

Find any stationary points of the graph of

1" title="y = 2x^2 + e^-x^4" alt="y = 2x^2 + e^-x^4" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
melomori [17]2 years ago
6 0

y = 2x {}^{2}  + e {}^{ - x {}^{4} }

\frac{dy}{dx}  = 4x +( e {}^{ - x {}^{4} } )( - 4x {}^{3} )

\frac{dy}{dx}  = 4x - 4x {}^{3} e { }^{ - x {}^{4} }

\frac{dy}{dx}  = 4x(1 - x {}^{2} e {}^{ - x {}^{4} } )

\frac{dy}{dx}  = 0 \\ 4x = 0 \:  \:  \:  \:  \: 1 - x {}^{2} e {}^{ - x {}^{4} }  = 0

4x = 0 \\ x = 0 \:   \:  \: \:  \:  \: y(0) = 0 + e {}^{0}  = 1

<h2>\frac{x {}^{2} }{e {}^{x {}^{4} } } = 1 \\ x {}^{2}  = e {}^{x {}^{4} }  \\ 2ln(x) = x {}^{4}  \\ these \: 2 \: functions \: do \: not \: intersect</h2><h2>Stationary point ( 0 , 1 )</h2>

<h2 />

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