Question

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">

Answer 1

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

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

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

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

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

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

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

Stationary point ( 0 , 1 )