What is |x^2-1|<= |x+1| ?

The solution to |x^2-1|<= |x+1| is x=-1\lor 0<= x<= 2

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|x^2-1|<= |x+1|

x^{2} - 1 \leq ∣ x + 1 ∣

x = - 1 or 0 \leq x \leq 2

Interval Notation

x = - 1 \cup [0, 2]

Solution steps

x^{2} - 1 \leq ∣ x + 1 ∣

Find positive and negative intervals

x \leq - 1, - 1 < x < 1, x \geq 1

Solve the inequality for each interval

x = - 1 or 0 \leq x < 1 or 1 \leq x \leq 2

Merge Overlapping Intervals

x = - 1 or 0 \leq x \leq 2

4 \cdot \frac{5}{6}

4 v < 8

y^{9}

15 < 4 + x