The range of the function g(x) = 2/(2 + 4x^2 + 5 + 9x^2) can be written as an interval (a,b]. What is a+b?
+9519 Simplify the function first.
\(g(x) = \dfrac2{2 + 4x^2 + 5 + 9x^2} = \dfrac2{13x^2 + 7}\)
Note that 13x^2 + 7 is always positive. So g(x) is always positive.
Also note that the minimum of 13x^2 + 7 is attained at x = 0, with a minimum value of 7.
Therefore, \(\displaystyle\max_{x\in\mathbb R} g(x) = \dfrac2{\displaystyle\min_{x\in\mathbb R}(13x^2 + 7)} = \dfrac27\).
Also, note that \(\displaystyle\lim_{x\to +\infty} g(x) = \displaystyle\lim_{x\to -\infty} g(x) = 0\).
Therefore, the range is \(\left(0, \dfrac27\right]\).
