depending on you level of math you solve this in different ways
calculus:
d/dx f(x)=5x^4-30x^2+9 so the function has maximum and minimums when the derivative equals 0 so 0=5x^4-30x^2+9 substituting instead of x^2 with u
0=5u^2-30u+9 using the quadratic formula and then taking the square root of the answers
x=-0.56278--->f(-0.56278)=-3.339 local minimum
x=0.56278----->f(0.56278)=3.339 local maximum
x=-2.3840---->f(-2.384)=37.030 global(on the interval) maximum
x=2.3840----->f(2.384)=-37.030 global(on the interval) minimum
you should note that this function does not have a global maximum or a minimum on the interval(-inf,inf).