The domain of the function q(x) = x^4 +4x^2 + 4 - 12x^2 - 18 is [0,inf). What is the range?
+9674 Note that:
\(\begin{array}{rcl} q(x) &=& x^4 + 4x^2 + 4 - 12x^2 - 18\\ &=& (x^4 + 4x^2 +4) - 12x^2 - 24 + 6\\ &=& (x^4 + 4x^2 +4) - 12(x^2 + 2) + 6\\ &=& (x^2 + 2)^2 -12(x^2 + 2) + 6\\ &=& (x^2 + 2)^2 - 12(x^2 + 2) + 36 - 30\\ &=& ((x^2 + 2)^2 - 12(x^2 + 2) + 36) - 30\\ &=& (x^2 - 4)^2 - 30 \end{array}\)
The minimum of q(x) is attained at x = 2, which is \(q(2) = (2^2 - 4)^2 - 30 = -30\).
Therefore, the range of q(x) in [0, inf) is \([-30, \infty)\).
+130071 Because of the leading x^4 term, the max is unbounded = inf
Using some Calculus to find the min
q'(x) = 4x^3 + 8x - 24x
q'(x) = 4x^3 - 16x
Set this = 0
4x^3 -16x = 0
4x (x^2 - 4) = 0
x= 0 or x =2
Taking the second derivative
12x - 16
When x = 0 this is negative .....so....not a min here
When x = 2 this is positive, so x is a min here
Petting 2 back into the function and evaluating gives us -30
So
Range = [ -30 , inf )
