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4 votes
4 votes
Find any stationary points of the graph of
y = 2x^2 + e^-x^4

User Tom Klino
by
2.8k points

1 Answer

15 votes
15 votes


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


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


(dy)/(dx) = 4x - 4x {}^(3) e { }^{ - x {}^(4) }


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


(dy)/(dx) = 0 \\ 4x = 0 \: \: \: \: \: 1 - x {}^(2) e {}^{ - x {}^(4) } = 0


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


\frac{x {}^(2) }{e {}^{x {}^(4) } } = 1 \\ x {}^(2) = e {}^{x {}^(4) } \\ 2ln(x) = x {}^(4) \\ these \: 2 \: functions \: do \: not \: intersect

Stationary point ( 0 , 1 )

User Aaru
by
3.2k points
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