Range of the function

Question

Let F(x)= $\frac{x^{2}-7x+10}{x^{2}-5x+6}$, then the range of f(x) is

(a). $\mathbb{R}$

(b)$\mathbb{R}-\{1\}$

(c) $\mathbb{R}-\{3\}$

(d) none of these

Answer Sheet Says (c) is correct.

$\boldsymbol{My}$ $\boldsymbol{Approach}$$\Longrightarrow$f(x)=$\frac{(x-2)(x-5)}{(x-3(x-2)}$ {Please tell me eliminating the factors from the numerator and denominator is a good idea or bad idea here ?}

$\Longrightarrow$$\frac{x-5}{x-3}$=1+$\frac{2}{3-x}$

x=2 $\Longrightarrow F(x)$=3 So book is wrong.So answer shoould be (a) $\mathbb{R}$

$\boldsymbol{Help}$$\boldsymbol{I}$ $\boldsymbol{Need}$$\Longrightarrow$Please tell me if book's answer is correct.

$\boldsymbol{I}$ $\boldsymbol{Need}$$\boldsymbol{Book}$$\boldsymbol{Suggestions}$ $\Longrightarrow$ Please suggest me books so that i can achieve a perfection in these kind of problems.

Answer 1

By your work we obtain:$$\mathbb R\setminus\{3,1\}.$$ Indeed, $\frac{x-5}{x-3}\neq1$ and $\frac{x-5}{x-3}\neq3$

Answer 2

Solve $$y=\frac{x^2-7 x+10}{x^2-5 x+6}$$ wrt $x\ne 2;\;x\ne 3$ you get $$x=\frac{3 y-5}{y-1}$$

$y\ne 1$ because this value makes zero the denominator

$y\ne 3$ because for $y=3$ would be $x=2$ that is excluded by the domain of the given function

$\dfrac{3 y-5}{y-1}=3$ has no solutions.

Thus the range is $\mathbb{R}\setminus\{1,3\}$

Answer 3

since $$x^2-5x+6=(x-2)(x-3)$$ you will have $$x\ne 2$$ and $$x\neq 3$$ since $$f(x)=\frac{(x-2)(x-5)}{(x-2)(x-3)}$$ and $$\lim_{x \to 2}f(x)$$ exists you can define $$f(x)=\frac{x-5}{x-3}$$ if $x\ne 2$ and $$f(x)=3$$ if $x=2$

Answer 4

x =/= 2 since this would make the denominator of the function zero in its original form. So it is true that 3 is not in the range. However, the book is wrong too, since 1 is also not in the range; indeed, if x satisfied f(x) = 1, the numerator and denominator in the original function would be equal, but solving here leads to x = 2.

Answer 5

$$f(x)=\dfrac{x^{2}-7x+10}{x^{2}-5x+6}=\dfrac{(x-2)(x-5)}{(x-2)(x-3)}\rightarrow Dom(f)=\mathbb{R}-\{ 2,3\}$$ If $x\neq 2\rightarrow f^{*}(x)=\dfrac{x-5}{x-3}=\dfrac{x-3-2}{x-3}=1-\dfrac{2}{x-3}$ $$f^{*}(2)=3; \dfrac{x^{2}-7x+10}{x^{2}-5x+6}=3\rightarrow x^{2}-7x+10=3x^{2}-15x+18\rightarrow$$ $$2x^{2}-8x+8=0\rightarrow x^{2}-4x+4=0\rightarrow x=2\notin Dom(f)\rightarrow \nexists x,\hspace{2mm} \text{for}\hspace{2mm} y=3$$ Also, $1-\dfrac{2}{x-3}=y\rightarrow \dfrac{2}{x-3}=1-y\rightarrow x-3=\dfrac{2}{1-y}\rightarrow \nexists x,\hspace{2mm} \text{for}\hspace{2mm} y=1$

As $\lim_{x\to 3^{-}}f(x)=+\infty\hspace{2mm} \wedge \lim_{x\to 3^{+}}f(x)=-\infty\rightarrow Rec(f)=\mathbb{R}-\{ 1,3\}$