If f(x)=(x2+2αx+α2−1)14 has its domain and range so that union is R, what does α satisfy?

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Makayla Santiago · Accepted Answer

Going from what you haveg(x)=(x−(−α+1))(x−(−α−1))≥0So, if you plot 1−α and −1−α on a number line; notice that the function g is positive everywhere to the right of (1−α) and also positive to the left of (−1−α) and thus satisfies the inequality whenx∈D=(−∞,−1−α]∪[1−α,∞)Moreover, notice that the vertex coordinates of the upward open parabola representing g is simply (−α,−1) (thereby making its range [−1,∞)). Therefore, the range of f, regardless of the value of α is R=[0,∞) (because of the fourth root on the outside).Now, we know D∪R=R⇒(−∞,−1−α]∪[1−α,∞)∪[0,∞)=RNotice that when α≤−1, we have the first interval in the above equality becomes (−∞,c] where c is non-negative, the second interval becomes redundant because of the third interval. Therefore, the inequality is held true. So α≥−1.

If f(x) = (x^2+2\alpha x + \alpha^2-1)^{\frac{1}{4}} has its domain

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