f = sqrt(x^4-3x^2+4)+sqrt(x^4-3x^2-8x+20)
To find the turning points, differentiate f and set that value to zero.
df/dx = (1/2)*(4*x^3-6*x)/sqrt(x^4-3*x^2+4)+(1/2)*(4*x^3-6*x-8)/sqrt(x^4-3*x^2-8*x+20) = 0
(4*x^3-6*x)/sqrt(x^4-3*x^2+4) = -(4*x^3-6*x-8)/sqrt(x^4-3*x^2-8*x+20)
cross-multiplying and squaring,
(4*x^3-6*x)^2*(x^4-3*x^2-8*x+20) = (4*x^3-6*x-8)^2*(x^4-3*x^2+4)
expanding this gives,
16*x^10-96*x^8-128*x^7+500*x^6+384*x^5-1068*x^4-288*x^3+720*x^2 = 16*x^10-96*x^8+244*x^6-236*x^4-64*x^7+288*x^5-544*x^3-48*x^2+384*x+256
which simplifies as,
64*x^7-256*x^6-96*x^5 +832*x^4-256*x^3-768*x^2+384*x+256 = 0
2x^7 - 8x^6 - 3x^5 + 26x^4 - 8x^3 - 24x^2 + 12x + 8 = 0
which factorises as,
(x^2 - 2)(2x^5 - 8x^4 + x^3 + 10x^2 - 6x - 4) = 0
Now the (x^2 - 2) term gives you the two roots, +/- sqrt(2)
The quintic equation has 3 real roots and two complex roots. The real roots are at (approx) x = -1, x = -0.5, x = 3.5. (I know where the roots are because I graphed the quintic and read from the graph.)
If you wish to get these roots more accurately, then use the Newton-Raphson method with the above root values as strarting values.
The fastest way of finding the minimum of the function, f, is to plug in all the above root values and determine the minimum value from that.
A faster method would be to graph the curve of f(x), then say, "From the graph of f(x) it can be seen that the smallest local minimum is at where x lies between x = 1 and x = 2. From the solution, x^2 - 2 = 0, we can say that the mimimum occurs at x = sqrt(2)."
Then include the minimum value.
f(sqrt(2)) = sqrt(4 - 6 + 4) + sqrt(4 - 6 - 8sqrt(2) + 20)
sqrt(2) + sqrt(18 - 8sqrt(2))
sqrt(2) + sqrt(16 - 8sqrt(2) + 2)
sqrt(2) + sqrt{(4 - sqrt(2))^2}
sqrt(2) + 4 - sqrt(2)
f(sqrt(2)) = 4
I'm afraid that I couldn't find all the real roots analytically, so this solution is a bit incomplete, however you may still find it useful.