Final answer:
To find the range of q(x) = x⁴ + 4x² + 4, we determine the minimum value of q(x) within the domain [0, ∞) and how it behaves as x increases. The minimum value is 4 and q(x) increases without bound, thus the range is [4, ∞).
Step-by-step explanation:
The domain of the function q(x) = x⁴ + 4x² + 4 is given as [0, ∞), which means that x takes on values from 0 to infinity. To find the range of the function, we need to find the minimum value of q(x) within this domain and see how it behaves as x increases towards infinity. Given that q(x) is a polynomial with even powers of x and positive coefficients, it has a minimum value when its derivative with respect to x is zero. Upon finding the derivative q'(x) = 4x³ + 8x, we set it to zero and solve for x to find the critical points. However, because the domain is restricted to non-negative x values, we can ignore any negative roots from the derivative. The only critical point occurs at x = 0. Evaluating q(x) at this point, q(0) = 0⁴ + 4*0² + 4 = 4, which is the minimum value of q(x) on the given domain. Since q(x) increases without bound as x approaches infinity, the range of q(x) on the domain [0, ∞) is [4, ∞).