Question

Calculate range of function:

$f(x) = \dfrac{(x+2)(x-1)}{x(x+1)}$

Answer 1

$\HugeR:(-\infty,1)\cup[9,\infty)$

$\Large\text{\underline{\underline{Important!}}}$

The x,y-plane only consists of real-valued coordinates.

$\Large\text{\underline{\underline{Explanation}}}$

$\text{ \rightarrow\boxed{f(x)=\dfrac{(x+2)(x-1)}{x(x+1)}} }$

$\cdots\longrightarrow x(x+1)f(x)=(x+2)(x-1)$

$\cdots\longrightarrow x^2f(x)+xf(x)-x^{2}-x+2=0$

$\text{ \cdots\longrightarrow \underline{\{f(x)-1\}x^{2}+\{f(x)-1\}x+2=0} .}$

We consider three eases for the quadratic equation in $x$.

Ⅰ. Discriminant of the equation is positive.

Ⅱ. Discriminant of the equation is zero.

Ⅲ. Discriminant of the equation is negative.

Each event results in the following cases.

Ⅰ. Two values of solutions to the equation.

Ⅱ. One value of the solution to the equation.

Ⅲ. No value of the solution to the equation.

Ⅲ gives no solution. Let's consider cases Ⅰ and Ⅱ only.

$\text{ \rightarrow\boxed{D=b^{2}-4ac\geq0, f(x)-1\neq0} }$

$\cdots\longrightarrow \{f(x)-1\}^{2}-8\{f(x)-1\}\geq0$

$\cdots\longrightarrow \{f(x)\}^{2}-2f(x)+1-8f(x)+8\geq0$

$\cdots\longrightarrow \{f(x)\}^{2}-10f(x)+9\geq0$

$\cdots\longrightarrow \{f(x)-9\}\{f(x)-1\}\geq0$

$\text{ \cdots\longrightarrow \underline{f(x)<1},\underline{f(x)\geq9} .}$

But, where the equation can be not quadratic,

$\rightarrow\boxed{f(x)-1=0}$

$\cdots\longrightarrow 0x^{2}+0x+2=0$

$\text{ \cdots\longrightarrow2=0 . (FALSE.)}$

Hence,

$\cdots\longrightarrow\boxed{R:(-\infty,1)\cup[9,\infty).}$

$\Large\text{\underline{\underline{More information}}}$

For anyone questioning, if we can narrow down the range even more, we can determine the exact range of the function by quadratic discriminant.

$\text{ D>0\iff The quadratic equation has two real distinct roots.}$

$\text{ D=0\iff The quadratic equation has two real identical roots.}$

$\text{ D<0\iff The quadratic equation has two imaginary distinct roots.}$

It is because 'if and only if($\iff$ in symbol)' shows two equivalent statements.

(e.g.) The quadratic equation(with real coefficients) can not have either two distinct real roots and $D\leq0$ simultaneously.

I hope all of you understood. Then, see you next time.