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whats the local minimum and local maximum of f(x)=x^4 - 6x^2 + 8
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local maximum: (0,8) local minima: (-1.732,-1) and (1.732,-1) Find these solutions by graphing and calculating the minimum and maximum.
or where f'(x)=0.
\[f(x)=x^4 - 6x^2 + 8 \]\[\implies f'(x)= 4x^3-12x = 4x(x^2-3) = 0 \implies x=0,\pm\sqrt{3} \]
\(f''(x)= 12x^2-12 < 0\) when x = 0 , maxima
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f′′(x)=12x2−12<0 when x = 0 , maxima I don't understand that
for maxima the sign of the 2nd derivative must change from positive to negative or negative to positive, it's the highest point on the graph besides the endpoints which increase without bound
ok thank you
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