The bottom line is: Take the derivative and set it equal to zero to find local minimums and then check the endpoints and find the one that has the lowest value.
For f(x) = 4sin2(x) + 5cos2(x) the derivative is perhaps easier to find if we write it like this:
f(x) = 4(sin(x))2 + 5(cos(x))2
To find the derivative, we need to use the power rule AND the chain rule so:
f '(x) = 8sin(x)cos(x) - 10sin(x)cos(x) = -2sin(x)cos(x)
Setting this equal to zero:
-2sin(x)cos(x) = 0 would occur if EITHER sin(x) = 0 or cos(x) = 0
sin(x) = 0 (on the interval [-π, π] occurs at x = -π, x = 0, and x = π
cos(x) = 0 (on [-π, π] occurs at x = -π/2 and x = π/2
So I can build a little table with these x values and plug them in to see wat f(x) is for each of them:
Notice I didn't need to separately add in the endpoints because they were critical points.
So the absolute minimum occurs at 2 points: x = -π/2 and x = π/2 and the value at both is 4
The answer to the question is 4 (it is the absolute minimum value)