Algebra> inequalities...

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

(x-2)² (x-3)²(x-4)²is 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.

shreyash kawalkar , 13 Years ago

Grade 11

3 Answers

Akash Kumar Dutta

we have.
(x-1)(x-3)/(x+1) ≥ 0
=>(x-1)(x-3)(x+1)≥ 0

=>plotting x=1,-1,3 and seeing the value of the above function by putiing values...

e.g.=>x=0 satisfies it so the interval between -1 and 1 is +
Hence the given domain of x=(-1,1] U [3,infinnte)
2 is the ans as already obtained at the first part.
to inlude it also in the the ans. a U {2} is added.
Regards.

Last Activity: 13 Years ago

Roshan Mohanty

2 is placed because at two f(x) assumes 0 na
and in ques it is given f(x) <= 0
dont forget that equal to sign

Now close brackets should be place when equal to sign is present
but take care not to place closed brackets there where function becomes undefined. ..

Last Activity: 13 Years ago