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find the stationary point on the curve y=3x^4 +1 and determine its nature

Sagot :

[tex]y=3x^4+1\\ y'=12x^3\\\\ 12x^3=0\\ x=0\leftarrow \text{stationary point}\\\\ \forall x\in(-\infty,0)\ y'<0\Rightarrow y\searrow\\ \forall x\in(0,\infty)\ y'>0\Rightarrow y\nearrow\\ \Downarrow\\ y(0)=y_{min}[/tex]

[tex]y_{min}=3\cdot0^4+1=1[/tex]
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