Let f: R - {2, 6} rightarrow R be a real-valu | vedclass

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(A) Let $y = \frac{x^2+2x+1}{x^2-8x+12}$.By cross-multiplying,we get:$y(x^2-8x+12) = x^2+2x+1$$yx^2 - 8xy + 12y = x^2 + 2x + 1$$x^2(y-1) - x(8y+2) + (

Vedclass · Accepted Answer

$\left(-\infty, -\frac{21}{4}\right] \cup [0, \infty)$

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(A) Let $y = \frac{x^2+2x+1}{x^2-8x+12}$.By cross-multiplying,we get:$y(x^2-8x+12) = x^2+2x+1$$yx^2 - 8xy + 12y = x^2 + 2x + 1$$x^2(y-1) - x(8y+2) + (12y-1) = 0$.Case $1$: If $y 
eq 1$,for $x$ to be real,the discriminant $D \geq 0$.$D = (-(8y+2))^2 - 4(y-1)(12y-1) \geq 0$$(64y^2 + 32y + 4) - 4(12y^2 - 13y + 1) \geq 0$$64y^2 + 32y + 4 - 48y^2 + 52y - 4 \geq 0$$16y^2 + 84y \geq 0$$4y(4y + 21) \geq 0$.This gives $y \in (-\infty, -\frac{21}{4}] \cup [0, \infty)$.Since $y 
eq 1$ was assumed,we check if $y=1$ is possible.Case $2$: If $y = 1$,then $x^2+2x+1 = x^2-8x+12$,which simplifies to $10x = 11$,or $x = \frac{11}{10}$.Since $x = \frac{11}{10}$ is in the domain $R - \{2, 6\}$,$y=1$ is included in the range.Thus,the range is $(-\infty, -\frac{21}{4}] \cup [0, \infty)$.

Let $f: R - \{2, 6\} \rightarrow R$ be a real-valued function defined as $f(x) = \frac{x^2+2x+1}{x^2-8x+12}$. Then the range of $f$ is

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