Question

what intervals on the graph f(x) = x^4 - 6x^2 convave upward? and which intervals concave downward? whic are inflection points?

Answer 1

sorry for the mis type...it should say cancave upward..

Answer 2

\[f \prime \left( x \right)=4x ^{3}-12x\]
\[f \prime \left( x \right)=0 gives 4x ^{3}-12x=0,4x \left( x ^{2}-3 \right)=0\]
\[x \left( x-\sqrt{3} \right)\left( x+\sqrt{3} \right)=0\]
\[x=0, \sqrt{3},-\sqrt{3}\]
\[f'\left( x \right)=4x \left( x-\sqrt{3} \right)\left( x+\sqrt{3} \right)\]
1. when x <0 slightly,
f'(x)=-(-)(+)=+ve
when x >0
f'(x)=+(-)(+)=-ve
f'(x) changes sign from +ve to -ve as x passes through 0
hence there is relative maxima at x=0
\[ 2. when x<\sqrt{3}slightly\],\[x-\sqrt{3}<0\]
f'(x)=+(-)(+)=-ve
\[when x> \sqrt{3},x-\sqrt{3}>0\]
f'(x)=+(+)(+)=+ve
therefore f'(x) changes sign from -ve to +ve as x passes through \[\sqrt{3}\]
Hence there is minima at x=sqrt3
\[3.when x <- \sqrt{3} slightly, x+\sqrt{3}<0\]
f'(x)=-(-)(-)=-ve
\[when x>-\sqrt{3},x+\sqrt{3}>0\]
f'(x)=-(-)(+)=+ve
Therefore f'(x) changes sign from -ve to +ve as x passes through \[-\sqrt{3}\]
Hence there is local minima at x=-sqrt3
at maxima concave downward.
At minima concave upward.