Question

Why is \[f(x)=|1-x+|x||\] continuous for every real \[x\]?

Options

• \[g(x)=1-x+|x|\] and \[h(x)=|x|\] are continuous, so \[f(x)=h(g(x))\] is continuous.

• \[f(x)\] is continuous only where \[\cos x\neq 0\].

• \[f(x)\] is continuous only when \[1-x+|x|=0\].

• \[f(x)\] is a rational function with \[q(x)\neq 0\].

Answer 1

Both the inner function \[g(x)=1-x+|x|\] and the outer function \[h(x)=|x|\] are continuous. By composition, \[f(x)=h(g(x))\] is continuous for every real \[x\].