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I know that the domain is $(0,1)$ since $x$ can't be $0$ or $1$. But I have tried to solve the range of this function without any success.

Can anyone please show me how I should think when solving a problem like this?

Thanks!

bsbb4
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FrustratedSpin
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3 Answers3

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In these types of questions it helps to analyze the function separately for disjoint intervals of the domain. Clearly, $x\ne0,1$ so the domain is $\Bbb R\setminus\{0,1\}$. That is to say, $x$ takes any real value but $0,1$.

For $x\in(-\infty,0)$, the function is continuous and differentiable. We need to find the minimum and maximum of $f$ in this interval, so equate the first derivative to $0$. This gives$$\frac1{x^2}=\frac4{(1-x)^2}\implies x=1/3,-1$$The point of interest is $-1$ at which $f$ obtains its maximum value of $1$ on $(-\infty,0)$. There is no minimum since $f(x)\to-\infty$ as $x\to0^-$.

Similarly analyze the function over $(0,1)$ and $(1,\infty)$. For the former, $f$ takes minimum value of $9$ at $x=1/3$ but no maximum exists since $f\to\infty$ as $x\to0^+$. For $(1,\infty)$, there is no minimum or maximum since $f\in(-\infty,0)$. Your range is the union$$(-\infty,1]\cup[9,\infty)\cup(-\infty,0)=(-\infty,1]\cup[9,\infty)=\Bbb R\setminus(1,9)$$

Shubham Johri
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For the case of real numbers, the range comes out to be as shown in the image :

enter image description here

As said in the comments section earlier, if $0 <x <1$, then you can directly apply Titu's lemma to get the answer as shown:

$1/x + 4/(1-x) \geq (1+2)^2 =9$

shsh23
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  • +1 Just wanted to say, it's preferable if you can format your answers with MathJax, without putting it into pictures. You seem capable of using MathJax, but here are a couple of tips: \implies gives $\implies$, \therefore gives $\therefore$, \cup give $\cup$, and \mathrm{Range} = gives $\mathrm{Range} = $. – Theo Bendit Sep 02 '19 at 10:53
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Let $f$ be given by $$ f(x) = \frac{1}{x} + \frac{4}{1-x} \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad\qquad $$

The domain of $f$ is the set of all real numbers other than $0$ or $1$.

To find the range of $f$ set $$ y=\frac{1}{x} + \frac{4}{1-x} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; $$ and solve for $x$ in terms of $y$.

\begin{align*} \text{Then}\;\;&y=\frac{1}{x} + \frac{4}{1-x}\\[4pt] \iff\;&yx^2+(3-y)x+1=0\;\;\text{and}\;\;x\not\in\{0,1\}\\[4pt] \iff\;&yx^2+(3-y)x+1=0\\[4pt] \iff\;& \left(y=0\;\,\text{and}\;\,x=-{\small{\frac{1}{3}}}\right) \;\;\text{or}\;\; \left( y\ne 0 \;\,\text{and}\;\, x=\frac {(y-3)\pm\sqrt{y^2-10y+9}}{2y} \right) \\[4pt] \end{align*} hence the only restriction on $y$ is the inequality $y^2-10y+9\ge 0$, or equivalently, $(y-1)(y-9)\ge 0$.

It follows that the range of $f$ is $(-\infty,1]\cup [9,\infty)$.

quasi
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