An algebra problem by manish kumar singh

Algebra Level 3

$f(x) = \begin{cases} 1 + x, x \geq 0 \\ 1 - x, x<0 \end{cases}$ Find $f(f(x))$ .

2-\mid x\mid

1+\mid x\mid

\mid2-\mid x\mid\mid

2+\mid x \mid

2 solutions

Tom Engelsman
Sep 18, 2017

The above function is described by $f(x) = |x| + 1, x \in \mathbb{R}$ . The composition $f(f(x))$ just equals $f(f(x)) = | |x|+1 | + 1 = (|x|+1) + 1 = \boxed{|x| + 2}.$

Manish Kumar Singh
Oct 16, 2015

By the definition of modulus function ,which states that :mod(x)=x, for x>0 or x=0 and mod(x)=-x, for x<0. we can write , f(x)=1+mod(x); hence f(f(x))= f(1+mod(x))=1+(1+mod(x))=2+mod(x).

An algebra problem by manish kumar singh

2 solutions

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