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Popular >

|s+1|>= 1

Solution

∣ s + 1 ∣ \geq 1

Solution

s \leq - 2 or s \geq 0

+1

Interval Notation

(- \infty, - 2] \cup [0, \infty)

Solution steps

∣ s + 1 ∣ \geq 1

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

or

u \geq a

s + 1 \leq - 1 or s + 1 \geq 1

s + 1 \leq - 1 or s + 1 \geq 1

s \leq - 2 or s \geq 0

Popular Examples

2 x^{2} + x \leq 1

∣ 3 x - 7 ∣ < 5

factor (4x+2)(4x-2)+6(4x+2)factor

(4 x + 2) (4 x - 2) + 6 (4 x + 2)

0 = - x^{2} + 4 x + 2

simplify ((x^2+y^2)^{1/4}+4(x^2+y^2)^{-3/4})/(x^2+y^2)simplify

\frac{( x ^{2} + y ^{2} ) ^{\frac{1}{4}} + 4 ( x ^{2} + y ^{2} ) ^{- \frac{3}{4}}}{x ^{2} + y ^{2}}

Frequently Asked Questions (FAQ)

What is |s+1|>= 1 ?
The solution to |s+1|>= 1 is s<=-2\lor s>= 0