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# Range

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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?

May 17, 2022

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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?

May 18, 2022