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How do you solve and graph #abs(x+1)<0#?

Answer 1

No solutions
graph{abs(x+1) [-10, 10, -5, 5]}

We can note that #|a| >= 0, |a|# is never negative for all real #a#, so hence #|x+1|# is never negative, hence no solutions to #|x+1|<0#.

To sketch this graph, we can consider #y = x+1# graph{y = x+1 [-10, 10, -5, 5]}

Then as we know #|a| >= 0# for all real a, we just take the graph in the domain where the function is negative, and reflect in the x axis, so for #x<-1# we make possitive, yielding;

graph{|x+1| [-10, 10, -5, 5]}

We can see clearly from this that #|x+1|# is never negative

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Answer 2

Genesis Allen

The absolute value of any real number is always non-negative, so the expression ( |x + 1| < 0 ) is never true for any real number ( x ). Therefore, there are no solutions to this inequality, and the graph of ( |x + 1| < 0 ) would be an empty set, represented by a number line with no points marked.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve and graph #abs(x+1)&lt;0#?

How do you solve and graph #abs(x+1)<0#?