I think my derivation is wrong because I can’t find the critical points (I’m getting complex numbers instead of real numbers) can you show me how to derive this and get critical points f(x)=2x-11-x²+2x2-x(1) dom f=\{2}asymptotes: lim f(x) = +os, lim f(x) = -0084488154(17) f(x) =I lim_ f(x) = +∞0₁2-22(x-1)-x²+2x2-x2-2+-x²+4x-22-22(-x+1)=x²+2x = -2²-22-x2-2-2²-22-2= -∞0 ⇒x=2 is a vertical asymptote-x² +42-2 → (-2x+4)(2-x) − (−x²+4x−2)(-1) = x²-4x+b⇒>>2-2(2-x)²(2-2)²(-22) (2-2) - (-2²-2)(-1)→>(2-2)²-x²-4x-2(2-2)²

Answered: Image /qna-images/answer/9548c0a7-de83-42c9-8f5a-a611048ff74b.jpg

Answered: I think my derivation is wrong because…

I think my derivation is wrong because I can’t find the critical points (I’m getting complex numbers instead of real numbers) can you show me how to derive this and get critical points

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

I think my derivation is wrong because I can’t find the critical points (I’m getting complex numbers instead of real numbers) can you show me how to derive this and get critical points

f(x)=2x-11-x²+2x
2-x
(1) dom f=\{2}
asymptotes: lim f(x) = +os, lim f(x) = -00
8448
8154
(17) f(x) =
I lim_ f(x) = +∞0₁
2-2
2(x-1)-x²+2x
2-x
2-2+
-x²+4x-2
2-2
2(-x+1)=x²+2x = -2²-2
2-x
2-2
-2²-2
2-2
= -∞0 ⇒x=2 is a vertical asymptote
-x² +42-2 → (-2x+4)(2-x) − (−x²+4x−2)(-1) = x²-4x+b
⇒>>
2-2
(2-x)²
(2-2)²
(-22) (2-2) - (-2²-2)(-1)
→>
(2-2)²
-
x²-4x-2
(2-2)²

Combination of a real number and an imaginary number. They are numbers of the form a + b , where a and b are real numbers and i is an imaginary unit. Complex numbers are an extended idea of one-dimensional number line to two-dimensional complex plane.

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