Show that the function $f$ defined by $f(x) = |1 - x + |x||$,where $x$ is any real number,is a continuous function.
(N/A) Let $g(x) = 1 - x + |x|$ and $h(x) = |x|$ for all real $x$.
Then the composite function $(h \circ g)(x) = h(g(x)) = h(1 - x + |x|) = |1 - x + |x|| = f(x)$.
Since $h(x) = |x|$ is a continuous function for all real $x$,and $g(x) = 1 - x + |x|$ is the sum of a polynomial function $(1 - x)$ and the modulus function $(|x|)$,both of which are continuous,$g(x)$ is also continuous.
Since $f(x)$ is the composition of two continuous functions $h$ and $g$,$f(x)$ is a continuous function for all real $x$.
Then the composite function $(h \circ g)(x) = h(g(x)) = h(1 - x + |x|) = |1 - x + |x|| = f(x)$.
Since $h(x) = |x|$ is a continuous function for all real $x$,and $g(x) = 1 - x + |x|$ is the sum of a polynomial function $(1 - x)$ and the modulus function $(|x|)$,both of which are continuous,$g(x)$ is also continuous.
Since $f(x)$ is the composition of two continuous functions $h$ and $g$,$f(x)$ is a continuous function for all real $x$.
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