The range of the function g(x) = 2/(2 + 4x^2) + 8/(1 + x^2)can be written as an interval (a,b]. What is a+b?
+9519 Note that\(g(x) = \dfrac2{2 + 4x^2} + \dfrac8{1 + x^2} = \dfrac1{1 + 2x^2} + \dfrac8{1 + x^2}\).
Also, \(x^2 + 1 \geq 1\) and \(1 +2x^2 \geq 1\).
Then, \(0 < \dfrac1{1 + 2x^2} \leq 1\) and \(0 < \dfrac1{1 + x^2} \leq 1\).
That means \(0 < \dfrac8{1 + x^2} \leq 8\).
Therefore, \(0 < g(x) \leq 9\) by adding the inequalities. Can you rewrite the range as an interval?
