how we solve |x+1|+|x-1| in term of function?
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here,
f(X) = |X+1|+|X-1|
when X=1
f(X) = 2
when X=-1
f(X)=2
here, we have use the property of modulous....
that, if f(X)= |X+a| then the value of X comes out to be -a.
Therefore the required function is f(X)=2
Hope you get your answer.
f(X) = |X+1|+|X-1|
when X=1
f(X) = 2
when X=-1
f(X)=2
here, we have use the property of modulous....
that, if f(X)= |X+a| then the value of X comes out to be -a.
Therefore the required function is f(X)=2
Hope you get your answer.
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