What is (2x^2-2)/(x^3-1)>= (x+1)/(x-1) ?

The solution to (2x^2-2)/(x^3-1)>= (x+1)/(x-1) is -1<= x<1

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(2x^2-2)/(x^3-1)>= (x+1)/(x-1)

\frac{2 x ^{2} - 2}{x ^{3} - 1} \geq \frac{x + 1}{x - 1}

- 1 \leq x < 1

Interval Notation

[- 1, 1)

Solution steps

\frac{2 x ^{2} - 2}{x ^{3} - 1} \geq \frac{x + 1}{x - 1}

Rewrite in standard form

\frac{- x ^{3} - 2 x - 3}{( x - 1 ) ( x ^{2} + x + 1 )} \geq 0

Factor

\frac{- x ^{3} - 2 x - 3}{( x - 1 ) ( x ^{2} + x + 1 )} : \frac{- ( x + 1 ) ( x ^{2} - x + 3 )}{( x - 1 ) ( x ^{2} + x + 1 )}

\frac{- ( x + 1 ) ( x ^{2} - x + 3 )}{( x - 1 ) ( x ^{2} + x + 1 )} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 1 ) ( x ^{2} - x + 3 ) ) ( - 1 )}{( x - 1 ) ( x ^{2} + x + 1 )} \leq 0 \cdot (- 1)

Simplify

\frac{( x + 1 ) ( x ^{2} - x + 3 )}{( x - 1 ) ( x ^{2} + x + 1 )} \leq 0

Identify the intervals

- 1 \leq x < 1

4 ∣ x - 3 ∣ - 8 = 8

b^{2} + 8 b + 16

- 5 u^{6} + 28 u^{5} - 15 u^{4} \leq 0

- 4

1 0^{- 9}

What is (2x^2-2)/(x^3-1)>= (x+1)/(x-1) ?
The solution to (2x^2-2)/(x^3-1)>= (x+1)/(x-1) is -1<= x<1