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Find out the range of $x^{\frac{1}{3}}+\sin 2x$


Since the range of the function $\sin2x$ is $[-1,1]$ and the range of the function $x^{\frac{1}{3}}$ is $R$.I found the range of the $x^{\frac{1}{3}}+\sin 2x$ as intersection of the ranges of the two functions $\sin2x$ and $x^{\frac{1}{3}}$ and i found the range as $[-1,1]$ but the range given in my book is $R$.

Just as the domain of sum/difference/product/quotient of two functions is the intersection of the domains of the individual functions.In the same way,i thought that the range of sum/difference/product/quotient of two functions is the intersection of the ranges of the individual functions.

But i am wrong.Is the range of sum/difference/product/quotient of two functions is the union/intersection of the ranges of the individual functions?

Or there is no such ''union/intersection'' relationship exist?Please explain.Thanks.

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No there is no direct "union/intersection" relationship.

Consider two functions with ranges that are a single interval, $[a,b]$ and $[c,d]$. Clearly, the sum of two values taken from these ranges cannot exceed $b+d$ nor be smaller than $a+c$, but there is no guarantee that these bounds can be reached, as the two function values do not vary independently.

As an extreme case, consider $f(x)=\sin(x)$ and $g(x)=-\sin(x)$, both with range $[-1,1]$. But $f(x)+g(x)=0$ !

All you can say a priori is that $\text{range}(f+g)\subseteq[a+c,b+d]$. More generally, $\text{range}(f+g)\subseteq\text{range}(f)\oplus\text{range}(g)$, where $\oplus$ denotes the Minkowski sum of the two sets.

To get the exact range, you need to perform a complete study, where knowing the ranges of the original functions helps little.

Similarly, $\text{range}(f-g)\subseteq[a-c,b-d]$ or $\subseteq\text{range}(f)\oplus\text{range}(-g)$. The cases of the product and quotient are a little more tricky...

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  • $\begingroup$ In the last line, why isn't the range given by $\text{range}(f-g) \subseteq [a - d, b - c ]$ ? $\endgroup$
    – DWade64
    Jul 31, 2018 at 14:57

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