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Regarding the function $f(x) = x\sqrt{3-x^2}$, I can determine that there are local min/max at +/-$\sqrt{\frac{3}{2}}$. I assumed these would also be the global max and min after looking at the graphed function. However, when I try to confirm my results via Wolfram Alpha, I am told that no global max or min exists. I can't seem to reason why. My best guess is that there is no global max/min because the function could go on and on either positively or negatively using imaginary numbers.

Is my guess correct? If not, why?

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  • $\begingroup$ It depends what you mean by "global". What definition of "global" are you using? $\endgroup$
    – Patrick Stevens
    Oct 3, 2015 at 15:44
  • $\begingroup$ The problem is from a calculus 1 class, so I assume the meaning is the largest value produced by the function. $\endgroup$
    – user184881
    Oct 3, 2015 at 15:45
  • $\begingroup$ And what is the domain of $f$? Remember that in general a definition of a function is not complete unless its domain and range are listed. $\endgroup$
    – Patrick Stevens
    Oct 3, 2015 at 15:46
  • $\begingroup$ Not specified, so I imagine the domain is over the reals. Not sure whether or not it is safe to assume the range is over the reals as well, but I suppose it would be consistent with the work we are doing in the class. $\endgroup$
    – user184881
    Oct 3, 2015 at 15:49
  • $\begingroup$ Your function is not defined over the whole real line. $\endgroup$
    – Clement C.
    Oct 3, 2015 at 15:52

4 Answers 4

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Let $f: \mathbb{R} \to \mathbb{C}$ by $x \mapsto x \sqrt{3 - x^2}$. Then you can easily show that $f$ takes real values only on $[-\sqrt{3}, \sqrt{3}]$, so it is quite inconvenient to ask for a "maximum" at all. (What is the "maximum" of $x \mapsto i x$, for instance?) You could insist on asking for the maximum and minimum modulus; the modulus is unbounded above as $x \to \infty$, and the minimum is zero (at $\pm \sqrt{3}$). In that sense, then, there are two global minima and no global maxima.

However, it would seem more sensible to let $f: [-\sqrt{3}, \sqrt{3}] \to \mathbb{R}$. Then $f$ takes real values everywhere, and it has a global maximum and a global minimum, at $\pm \sqrt{3}$ respectively. (Here, I interpret "global" to mean "on the entire domain".)

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prove that $$x\sqrt{3-x^2}\le \frac{3}{2}$$ iff $$0\le \left(x^2-\frac{3}{2}\right)^2$$

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A function is well defined only if we specify the domain and codomain. For $y=x\sqrt{3-x^2}$ we can define the function: $$ f: [-\sqrt{3},\sqrt{3}]\rightarrow \mathbb{R} $$ that has alocal and global maximum at $\sqrt{3/2}$ and a local and global minimum at $\sqrt{3/2}$.

But the function: $$ f:\mathbb{R} \rightarrow \mathbb{C} $$ has no global maxima.

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Hint: $x\sqrt{3-x^2}$ is an odd function on $\left[-\sqrt3,\sqrt3\right]$, so its maximum will be the negative of its minimum. The square of its maximum will be the maximum of $$ x^2\left(3-x^2\right)=\frac94-\left(x^2-\frac32\right)^2 $$

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