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How to find range of $$f(x) = x^2 + \frac{1}{x^2+1} \quad ?$$

Its domain is all real numbers. If I use calculus it is very lengthy but if I put RHS of equation equal to y I get a quadratic equation in $x^2$ but I cannot find range of y by imposing condition on discriminant as even if let us say $x$ was complex or imaginary but there is the possibility that when it is raised to power 2 or 4 it becomes purely real?

Matt
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3 Answers3

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Hint: write it as $\,(x^2+1)+\cfrac{1}{x^2+1}-1\,$ and use $\,a+\cfrac{1}{a}\ge 2\,$ for $\,a \ge 1\,$.

dxiv
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    Wow ! Cant believe it could be done so easily . Just what I was looking for ! Giving you a plus one. – Matt May 08 '17 at 15:26
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Clearly the range is all positive as both terms are positive. As $x$ gets large it goes to $+\infty$ so you just need to find the minimum. Take the derivative, set to zero, find the minimum, and you are done.

Ross Millikan
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First notice that the function is even, that is $f(x) = f(-x)$.

I'd like to show that $1$ is either the upper or lower bound of the function. We have $f(0) = 1$, and solving the equation $f(x) = 1$ gives only one solution. Assume there are two real numbers $a$ and $b$ such that $f(a) < 1 < f(b)$. If any of them are negative, simply use $f(x) = f(-x)$ and change their sign, thereby assuming WLOG that they are both positive. By the intermediate value theorem there must be a $c$ between $a$ and $b$ such that $f(c) = 1$, but since $0 < c$ this is a contradiction. Therefore all real numbers in the range must be to the same side of $1$.

Since $x^2 < f(x)$ and $x^2 \to \infty$ as $x \to \infty$ we must have $f(x) \to \infty$ as $x \to \infty$.

From this we can conclude that the range of the function is $[1;\infty[$.

Alice Ryhl
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  • Your argumentation based on the number of preimages (Card $f^{-1}(1)=1$ and not an even number, thus...) is "borderline". You should make appeal to classical analysis theorems... – Jean Marie May 08 '17 at 15:36
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    @JeanMarie It's pretty easy to make that argument rigorous: Assume there are two real numbers $a$ and $b$ such that $f(a) < 1 < f(b)$, since the function is symmetric around the $y$-axis, we have $f(x) = f(-x)$ so if any of $a,b$ are negative, we can WLOG change their sign. By the intermediate value theorem we have some value $c$ such that $a < c < b$ or $b < c < a$ and $f(c) = 1$, but since $0 < c$ and the only solution to $f(x) = 1$ is $x=0$ we have a contradiction. – Alice Ryhl May 08 '17 at 16:25
  • Thanks : this is a neat improvement. Btw, a liitle detail, it is better to say that a function is even than symmetric around the $y$-axis (which is property of its graphical representation) – Jean Marie May 08 '17 at 18:23
  • -1 The fact that a function is symmetric around the $y$-axis doesn't mean it has a local min or max at 0. Even adding continuous and differentiable doesn't guarantee that. See $x^4(\cos(1/x)$ with the obvious completion of $0$ at $x=0$. – DRF May 09 '17 at 04:39
  • @DRF Okay, so you are right. Luckily with the argument given in the previous comment, we don't really need that there is a local min or max at 0. – Alice Ryhl May 09 '17 at 05:32