We have the function \begin{equation*}f(x)=-\frac{(x-3)^2}{x+1}\end{equation*}

I want to determine the domain and the range of the function.

The root of the deniminator is $x=-1$. Therefore, the domain is $D_f=\mathbb{R}\setminus\{-1\}=(-\infty, -1)\cup (-1,+\infty )$.

Does it holds that \begin{equation*}W_f=f(D_f)=f\left ((-\infty, -1)\cup (-1,+\infty )\right )=f\left ((-\infty , -5)\cup (-5,-1) \cup (-1,3) \cup (3,+\infty)\right )\end{equation*} ?

If this is correct, then we have to know the monotonicity of $f$ at each of these intervalls.

We have that at $(-\infty , -5)$ the function is decreasing, at $(-5,-1)$ the function is incresing, at $(-1,3)$ the function is increasing and at (3,+\infty)$ the function is decreasing.

We have the following:

  • \begin{align*}f\left ((-\infty , -5)\right )= \left (16, +\infty\right ) \end{align*}

  • \begin{align*}f\left ((-5,-1)\right )= \left (16, +\infty\right )\end{align*}

  • \begin{align*}f\left ((-1,3)\right )= \left (-\infty, 0\right )\end{align*}

  • \begin{align*}f\left ((3,+\infty)\right ) = \left (-\infty, 0\right )\end{align*}

Therefore the range is \begin{equation*}W_f= \left (-\infty, 0\right )\cup \left (16, +\infty\right ) \end{equation*}


Is everything correct? Or do we have to determine the range in an other way?

  • 2
    $\begingroup$ You should be including 0 and 16 in the range $\endgroup$ – David Quinn Sep 22 '17 at 12:00
  • 1
    $\begingroup$ Oh yes... Because we write the interval $(-\infty, -1)$ as $(-\infty, -5]\cup [-5, -1)$ and the interval $(-1, +\infty)$ as $(-1,3]\cup [3,+\infty)$, right? @DavidQuinn $\endgroup$ – Mary Star Sep 22 '17 at 12:08


Thus, for $x<-1$ by AM-GM we obtain $$f(x)=8-x-1-\frac{16}{x+1}\geq8+2\sqrt{(-x-1)\left(-\frac{16}{x+1}\right)}=16$$ and for $x>-1$ by AM-GM again we obtain: $$f(x)=8-x-1-\frac{16}{x+1}\leq8-2\sqrt{(x+1)\cdot\frac{16}{x+1}}=0.$$ Since $\lim\limits_{x\rightarrow-1^+}f(x)=-\infty$ and $\lim\limits_{x\rightarrow-1^-}f(x)=+\infty$ and $f$ is a continuous function

on all interval of the domain, we got the answer: $$(-\infty,0]\cup[16,+\infty).$$


You do not "have to" do it another way, but you can do it algebraically:

Say $y = f(x)$, then solve $$y = - \frac{(x-3)^2}{x+1}$$ for $x$. What you get is $$x = 3 - \frac{y}{2} \pm \frac{1}{2} \sqrt{y^2 - 16y} \;.$$ For a solution in the real numbers, you need $$y^2 - 16y \ge 0 \iff y \le 0 \wedge y \ge 16 \;.$$ You still have to check that against the domain that you determined earlier, so I am not sure which is less work in the end.

  • $\begingroup$ So, can we find the inverse $f^{-1}$ and find the domain of $f^{-1}$ which is equal to the range of $f$ ? $\endgroup$ – Mary Star Sep 24 '17 at 7:39
  • $\begingroup$ You use that $f$ is surjective on its image: For every element $y$ in the image of $f$, there must be an element $x$ that maps to it: $y \in \mathop{\mathrm{im}} f \iff \exists x \in \mathop{\mathrm{dom}} f : f(x) = y$. However, care must be taken to only consider those solutions $x$ in the domain of the function if it is somehow restricted. $\endgroup$ – user480945 Sep 24 '17 at 11:05

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