#f(x)=12+2x^2-4x^4#
i guess that is your function
so, first of all, find its first derivative and that is
#f'(x)=4x-16x^3#
now solve for #x# by making #f'(x)=0#
#4x-16x^3=0#
#4x(1-4x^2)=0#
#4x=0# therefore #color(red)(x=0#
#1-4x^2=0# therefore #color(red)(x=1/2#
now find the second derivative and that is
#f''(x)=4-48x^2#
now substitute the values of #x# in the original function to find the exact minima and maxima points
when #x=0#
#f(0)=12+2(0)^2-4(0)^4#
#f(0)=12#,
#therefore(0,12)#
when #x=1/2#
#f(1/2)=12+2(1/2)^2-4(1/2)^4#
#f(1/2)=49/4#,
#therefore(1/2,49/4)#
now we have the points #(0,12)# and #(1/2,49/4)#
now substitute the values of #x# in the second derivative to find the nature of stationary points
when #x=0#
#f''(0)=4-48(0)^2#
#f''(0)=4# which is #>0#
#therefore# we have a local minima at #x=0# and that is #(0,12)#
when #x=1/2#
#f''(1/2)=4-48(1/2)^2#
#f''(1/2)=-8# which is #<0#
#therefore#we have a local maxima at #x=1/2# ansd that is #(1/2,49/4)#
now we can say that there is a local minima at #(0,12)# and local maxima at #(1/2,49/4)#
NOTE that :
For a function #f(x)# with a stationary point #x=a#
. If #f''(a)>0#, the it is a local minima
. If #f''(a)<0#, then it is a local maxima
. If #f''(a)=0#, then it could be a local maxima, a local minima or a stationary inflection point
i hope that helps
