Please explain in brief the wavy curve method & also solve this ques..

f(x)=(x-2)²(1-x)(x-3)³(x-4)²/(x+1) ≤ 0 find the value of X.

Dear neeraj

f(x)=(x-2)²(1-x)(x-3)³(x-4)²/(x+1) ≤ 0

 (x-2)² (x-3)²(x-4)²   is always positive

at x=2,3,4   (x-2)² (x-3)²(x-4)² is  zero 

so x=2,3,4 is a solution  

now for  (1-x)(x-3)/(x+1) ≤ 0

  or        (x-1)(x-3)/(x+1) ≥ 0

so silution is [-1,1] and [3,∞) 

so final solution   [-1,1] and [3,∞)  and 2

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Badiuddin

(-1,1] U [3,∞) U {2}      

sir,i am unable to understand the solution for the question.please explain clearly

-(x-2)² (x-1)(x-3)³ (x-4)² /(x+1)≤0

(x-2)² (x-1)(x-3)³ (x-4)² /(x+1)≥0

equate every linear factor by 0,

x=1,2,3,4,-1

by ploting the values nof x on no line we get,

x Ε (-∞,-1]υ[1,2]υ[3,4]υ[4,∞)

........

HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.

AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.

PLEASE HELP ME. FAST.

f(x)=(x-2)²(1-x)(x-3)³(x-4)²/(x+1) ≤ 0

 (x-2)² (x-3)²(x-4)²   is always positive

at x=2,3,4   (x-2)² (x-3)²(x-4)² is  zero 

so x=2,3,4 is a solution   

now for  (1-x)(x-3)/(x+1) ≤ 0

  or         (x-1)(x-3)/(x+1) ≥ 0

ans:   (-1,1] U [3,∞) U {2}

HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.

AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.

PLEASE HELP ME. FAST.

explain to me: why does the wavy curve method hold?  what is the actual idea behind it?