Given \( f(x) = \begin{cases} 1 & \text{if } |x| \le 1 \\ 0 & \text{if } |x| > 1 \end{cases} \) and \( g(x) = \begin{cases} 2-x^2 & \text{if } |x| \le 2 \\ 2 & \text{if } |x| > 2 \end{cases} \). If \( h(x)=f[g(x)] \) then an interval in which \( h(x)=1 \) for all values of \( x \) in that interval is
Step-by-step Solution:
We need to find the interval where \( h(x) = f(g(x)) = 1 \).\nFrom the definition of \( f(x) \), \( f(y) = 1 \) if and only if \( |y| \le 1 \).\nThus, we require \( |g(x)| \le 1 \).\n\nCase 1: Let \( |x| > 2 \).\nIn this case, \( g(x) = 2 \). We check if \( |g(x)| \le 1 \).\n\( |2| \le 1 \) is False.\nSo no solution in this interval.\n\nCase 2: Let \( |x| \le 2 \).\nIn this case, \( g(x) = 2 - x^2 \). We require \( |2 - x^2| \le 1 \).\nThis splits into an inequality:\n\[ -1 \le 2 - x^2 \le 1 \]\nSubtract 2 from all sides:\n\[ -3 \le -x^2 \le -1 \]\nMultiply by -1 (reversing the inequalities):\n\[ 1 \le x^2 \le 3 \]\nTaking the square root gives:\n\[ 1 \le |x| \le \sqrt{3} \]\n\nThis satisfies the initial condition \( |x| \le 2 \) (since \( \sqrt{3} < 2 \)).\nThus, the required interval is \( 1 \le |x| \le \sqrt{3} \).