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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→((x^2 +x−1)/(x−1)) where x≠1 find the range of the function
f:x→x2+x−1x−1wherex≠1findtherangeofthefunction
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)=x2+x−1x−1=x+2+1x−1limx→±∞f(x)=±∞,limf(x)=±∞x→1±f′(x)=1−1(x−1)2f′(x)=0⇔x=2f(0)=1andf(2)=5∀x∈]0,1[∪]1,2[,f′(x)<0∀x∈]−∞,0[∪]2,+∞[,f′(x)>0f(]−∞,0])=]−∞,1]f([0,1[)=[−∞,1]f(]1,2])=[5,+∞]f([2,+∞[)=[5,+∞]∙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)=x2+x−1⇒x2+(1−y)x+y−1=0⇒Δ⩾0⇉(1−y)2−4(y−1)⩾0⇉(y−1)(y−5)⩾0⇉y⩽1∪y⩾5Rf={y∈R∣y⩽1∪y⩾5}

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