What is the local minimum value of the function? (Round answer to the nearest hundredth) g (x) = x^4-5x^2+4

The local minimum of function is an argument x for which the first derivative of function g (x) is equal to zero, so:

g' (x) = 0

g' (x) = (x^4-5x^2+4) '=4x^3-10x=0

x (4x^2-10) = 0

x=0 or 4x^2-10=0

4x^2-10=0 / 4

x^2-10/4=0

x^2-5/2=0

[x-sqrt (5/2) ][x+sqrt (5/2) ]=0

Now we have to check wchich argument gives the minimum value from x=0, x=sqrt (5/2) and x=-sqrt (5/2).

g (0) = 4

g (sqrt (5/2)) = 25/4-5*5/2+4=4-25/4=-9/4

g (-sqrt (5/2)) = - 9/4

The answer is sqrt (5/2) and - sqrt (5/2).