Question 997840
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f(x) = {{{x^4 + 2x^2 - 3}}}.


Take the derivative:


f'(x) = {{{4x^3 + 4x}}}.


Find where the derivative is zero:   {{{4x^3 + 4x}}} = {{{0}}}.


The only real root is  x = 0.


Calculate f(0).  It is  -3.


The values at the ends  x=-1  and  x=2  are greater.


So,  the minimum is at  x=0.  The maximum is at  x=2.