SOLUTION: Simplify the inequality, identify any critical points, and graph its solution
{{{1-(2x^"")/(x^2+1)-(1+4x-3x^2)/(x^3-2x^2+x-2)>-1^""/(x^""-2) }}}
https://ibb.co/eaRKga
(th
Algebra ->
Rational-functions
-> SOLUTION: Simplify the inequality, identify any critical points, and graph its solution
{{{1-(2x^"")/(x^2+1)-(1+4x-3x^2)/(x^3-2x^2+x-2)>-1^""/(x^""-2) }}}
https://ibb.co/eaRKga
(th
Log On
We factor the denominator
Factor out x2 from the first two terms and
factor out +1 from the last two terms:
Factor out (x-2)
Write 1 as
The LCD is (x-2)(x²+1)
Multiply each numerator and denominator by whatever
factor(s) that are needed so the resulting denominator
will become the LCD.
Combine all the numerators over the LCD:
Combining terms:
That numerator x3+x-2 obviously has zero 1.
So we factor it using synthetic division:
1 | 1 0 1 -2
| 1 1 2
1 1 2 0
So we see that it factors as (x-1)(x2+x+2)
The critical numbers are real zeros of the numerator and
zeros of the denominator.
The only real zero of the numerator is 1
The only real zero of the denominator is 2
We place those critical numbers on a number line
They cannot be solutions themselves because the
inequality does not permit equality:
-----------o----o----------
-1 0 1 2 3 4
We choose a test value in the interval left of 1
The easiest one is 0 and substitute it into the
inequality:
That is true, so we shade the number line left of 1
<==========o----o----------
-1 0 1 2 3 4
We choose a test value between 1 and 2. The easiest
one is 1.5 and substitute it into the
inequality:
That is false, so we do not shade the number line between
1 and 2. So we still have this graph:
<==========o----o----------
-1 0 1 2 3 4
We choose a test value in the interval right of 2
The easiest one is 3 and substitute it into the
inequality:
That is true, so we shade the number line right of 2
<==========o----o==========>
-1 0 1 2 3 4
That is the graph of the solution. In interval notation
that is written:
Edwin