f:x→((x^2 +x−1)/(x−1)) where x≠1 find the range of the function


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Question Number 148241 by 7770 last updated on 26/Jul/21

$f : x \to \frac{x^{2} + x - 1}{x - 1} w h e r e x \neq 1$ $f i n d t h e r a n g e o f t h e f u n c t i o n$
	Answered by Olaf_Thorendsen last updated on 26/Jul/21
	$f(x) = ((x^2 +x−1)/(x−1)) = x+2+(1/(x−1)) lim_(x→±∞) f(x) = ±∞, limf(x) = ±∞_(x→1^± ) f′(x) = 1−(1/((x−1)^2 )) f′(x) = 0 ⇔ x = 2 f(0) = 1 and f(2) = 5 ∀x∈]0,1[∪]1,2[, f′(x) < 0 ∀x∈]−∞,0[∪]2,+∞[, f′(x) > 0 f(]−∞,0]) = ]−∞,1] f([0,1[) = [−∞,1] f(]1,2]) = [5,+∞] f([2,+∞[) = [5,+∞] • f(R\{1}) = R\]1,5[$
	$f (x) = \frac{x^{2} + x - 1}{x - 1} = x + 2 + \frac{1}{x - 1}$ $\lim_{x \to \pm \infty} f (x) = \pm \infty, \underset{x \to 1^{\pm}}{\lim f (x) = \pm \infty}$ $f^{'} (x) = 1 - \frac{1}{{(x - 1)}^{2}}$ $f^{'} (x) = 0 \Leftrightarrow x = 2$ $f (0) = 1 and f (2) = 5$ $\forall x \in] 0, 1 [\cup] 1, 2 [, f^{'} (x) < 0$ $\forall x \in] - \infty, 0 [\cup] 2, + \infty [, f^{'} (x) > 0$ $f (] - \infty, 0]) =] - \infty, 1]$ $f ([0, 1 [) = [- \infty, 1]$ $f (] 1, 2]) = [5, + \infty]$ $f ([2, + \infty [) = [5, + \infty]$ $∙ f (R ∖ {1}) = R ∖] 1, 5 [$
	Answered by liberty last updated on 26/Jul/21
	$y(x−1)=x^2 +x−1 ⇒x^2 +(1−y)x+y−1=0 ⇒Δ≥0 ⇉ (1−y)^2 −4(y−1)≥0 ⇉ (y−1)(y−5)≥0 ⇉ y≤1 ∪ y≥ 5 R_f = {y∈R∣ y≤1 ∪ y≥ 5 }$
	$y (x - 1) = x^{2} + x - 1$ $\Rightarrow x^{2} + (1 - y) x + y - 1 = 0$ $\Rightarrow Δ ⩾ 0$ $⇉ {(1 - y)}^{2} - 4 (y - 1) ⩾ 0$ $⇉ (y - 1) (y - 5) ⩾ 0$ $⇉ y ⩽ 1 \cup y ⩾ 5$ $R_{f} = {y \in R ∣ y ⩽ 1 \cup y ⩾ 5}$
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