total mod |x^2-5x+4/(x^2-4)|< or equal to 1

[half swift]

total mod |x^2-5x+4/(x^2-4)|< or equal to 1

[old vortex]

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[nova mist]

so okay for writing sake I will say f(x) is the inside of the absolute value part

[half swift]

Ok

[nova mist]

so we can drop the absolute value easily

by doing

-1 < f(x) < 1

all these values will also hold for |f(x)| < 1

[half swift]

Ok so i did the solving and got 2 cases

so do we need to do intersection or union for these two cases?

Thats my doubt

[nova mist]

yeah what I would do is find when it is greater than -1

then when is it less than 1

and find where both are true

[half swift]

if we solve fx<1 then we get x belongs to [-2,8/5]U[2,infinity)

[nova mist]

okay now solve for -1 < f(x)

[half swift]

and for -1 we get x belongs to (-infinity , -2]U[0,2]U[5/2 , infinity)

so do i do union or intersection?

for these 2

(-infinity , -2]U[0,2]U[5/2 , infinity) and [-2,8/5]U[2,infinity)

[nova mist]

not union but the intersection

when are both true

when is it greater than -1 but less than 1

[half swift]

why intersection?

[nova mist]

because that is where they are both true iirc

at the intersection both should be true

[half swift]

what does both are true mean?

Like both cases are possible

[nova mist]

so we have 2 cases that must be true that is

-1 < f(x) and f(x) < 1 for |f(x)} < 1

[half swift]

Ye

[nova mist]

so for this you need to find when f(x) < 1

once you find that find when -1 < f(x)

we know that stuff good

[half swift]

Ok

[nova mist]

now when both are true so when f(x) is less than 1 but greater than -1

the inequality we need is true

so find when that is the case

[half swift]

Oh ok so intersection

[nova mist]

yes

[half swift]

only if -a<x<a?

[nova mist]

yep and for our case a = 1

[half swift]

Ok

thanks

[nova mist]

so we have all our cases

[half swift]

Ok

[nova mist]

if you send me the answer when you get it I can fact check it

[half swift]

Ye ik the answer

just wanted to know why intersection here

Ok thanks

+close