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I have this function:

$$f(x)=\frac{x}{x^2+1}$$

it's clear that the domain is all real numbers, because the denominator never become zero. I've done the following procedure to found the range of $f(x)$:

$\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \operatorname{Dom} f,y=f(x)} \right \} \\ \operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},y=\dfrac{x}{x^2+1}} \right \} \\ \operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},x=\dfrac{1\pm \sqrt{1-4y}}{2y}} \right \} \\ \operatorname{Rec} f=\left \{{y\in \mathbb{R}:\left (1-4y^2 \right )\geq 0\, \wedge\, y\neq 0 } \right \}\\ \operatorname{Rec} f=[-\tfrac{1}{2},\tfrac{1}{2}]-\left \{ 0 \right \}$

But, the function is zero when $x=0$ $(f(0)=0)$, and the range of the function is $[-\tfrac{1}{2}, \tfrac{1}{2}]$. So I don't know where is my error in the procedure.

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You have the equation $y = \frac{x}{x^2 + 1}$, which you then simplified to $x^2y + y - x = 0$. Then, solving this for $x$ gives you $\frac{1 \pm \sqrt{1-4y^2}}{2}$, which is all good, very good, until...

If $y = 0$, then the above is not even a quadratic equation, so you cannot approach it through the quadratic formula! Instead, you must deal with such a case separately, by setting $y = 0$, which gives $x = 0$.

Moral of the (real) story : The quadratic formula applies only when the quadratic coefficient is non-zero. Otherwise, such a quadratic simplifies to a linear (possibly, even better) equation which can be solved.

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  • $\begingroup$ When I plot the graph $x = \frac{1\pm\sqrt{1-4y}}{2y}$, I get something completely different than $y = \frac{x}{x^2+1}$. Could this be part of it? $\endgroup$ – rb612 Apr 27 '18 at 0:19
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    $\begingroup$ You mean $\sqrt{1-4y^2}$? You've made this typo a few places. $\endgroup$ – Ted Shifrin Apr 27 '18 at 0:23
  • $\begingroup$ Correct, the mistake propagated into the answer as well. $\endgroup$ – астон вілла олоф мэллбэрг Apr 27 '18 at 0:24
  • $\begingroup$ @астонвіллаолофмэллбэрг You really should not upvote questions as a reward for accepting your answer :/ $\endgroup$ – Ant Apr 27 '18 at 8:49
  • $\begingroup$ @Ant You are correct. However, I have upvoted this question because of its quality, not because I have answered it. Usually, when I upvote or downvote I mention it. In this case since I mentioned it under the answer the wrong message would have been sent. I apologize for that : I have only upvoted for the quality of the question, not because I answered it. Having said that , I don't even answer questions that don't have good quality in my opinion, so if I've answered it that means it has some quality, which I appreciate via upvotes etc. $\endgroup$ – астон вілла олоф мэллбэрг Apr 27 '18 at 9:11
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Note that $f(0) = 0$ and $f(x)\to0$ as $x\to\infty$ and $f(x)>0$ when $x>0,$ and $f$ is everywhere continuous. That implies that as $x$ goes from $0$ to $\infty,$ then $f(x)$ must go up from $0$ and ultimately back down to $0.$ Thus $f$ has an absolute maximum somewhere to the right of $0.$ And $f(x)$ must assume all intermediate values between $0$ and that maximum.

Next: $$ \frac x {x^2+1} = \overbrace{ \frac 1 {x + \frac 1 x} = \frac 1 {\left( x - 2 + \frac 1 x \right) + 2} }^\text{This is a sort of completion of the square} = \frac 1 {\left( \sqrt x - \frac 1 {\sqrt x} \right)^2 + 2}. $$ This is equal to $1/2$ when the square is $0;$ otherwise it is less than $1/2.$

Thus $f$ maps $[0,\infty)$ to $[0,1/2].$

Since $f$ is an odd function, it therefore maps $(-\infty,0]$ to $[-1/2,0].$

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You lost the point $0$ from the range when you applied the quadratic formula. It implicitly assumes that you're dealing with a quadratic. The solution to $$ax^2 + bx + c = 0$$ is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$ provided $a \neq 0$. If $a = 0$, then this equation becomes linear, and the quadratic formula becomes nonsense.

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You have that $\operatorname{Ran} f= \{0\}\cup\{y: y\neq 0\}$. What you did was to solve for the case $y \neq 0$. Every thing else is fine with your solution.

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$$f(x)=\frac{x}{x^2+1}$$

Since the function is an odd function $(f(x)=-f(-x))$, then the minimum value is equal to the opposite of the maximum value. We see that $f(x) \ge 0$ when $x \ge 0$, so we need to find the maximum value of $f(x)$ when $x \ge 0$.

Note that $\dfrac 12 - f(x) = \dfrac 12 - \dfrac{x}{x^2+1} = \dfrac{x^2+1-2x}{2(x^2+1)} = \dfrac{(x-1)^2}{2(x^2+1)}$

So, when $x \ge 0$, we see that $\dfrac 12 - f(x) \ge 0$ and, when $x=1$, we see that $\dfrac 12 - f(x) = 0$.

Since $\dfrac 12 - f(x) \ge 0 \iff f(x) \le \dfrac 12$ and $f(1)= \dfrac 12$, we see that $\displaystyle \max_{x \ge 0} f(x) = \dfrac 12$. Hence, because $f(x)$ is an odd function, $\operatorname{Range} f(x) = \left[ -\dfrac 12, \dfrac 12 \right]$

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