Let \( f(x)=2 x+1, g(x)=-x^{2}+1 \), and \( h(x)=\frac{1}{x+1} \) The function \( t \) is defined as \( t(x)=\left\{\begin{array}{ll}h(x) & \text { if } x \in(2,6] \\ g(x) & \text { if } x \leq 1\end{array}\right. \) The domain of function \( t \) is:

To determine the domain of the function \( t(x) \), we need to analyze the conditions under which each piece of the piecewise function is defined. 1. **For \( h(x) = \frac{1}{x+1} \)**: - The function \( h(x) \) is defined for all \( x \) except where the denominator is zero. Thus, \( h(x) \) is undefined at \( x = -1 \). - The piece \( h(x) \) is used for \( x \in (2, 6] \). Therefore, within this interval, \( h(x) \) is defined. 2. **For \( g(x) = -x^2 + 1 \)**: - The function \( g(x) \) is a polynomial and is defined for all real numbers. - The piece \( g(x) \) is used for \( x \leq 1 \). Therefore, \( g(x) \) is defined for all \( x \) in this range. 3. **Combining the intervals**: - The function \( t(x) \) is defined as: - \( t(x) = h(x) \) for \( x \in (2, 6] \) - \( t(x) = g(x) \) for \( x \leq 1 \) Now, we need to combine these intervals to find the overall domain of \( t(x) \): - From \( g(x) \), we have the interval \( (-\infty, 1] \). - From \( h(x) \), we have the interval \( (2, 6] \). Since there is a gap between \( 1 \) and \( 2 \), the overall domain of \( t(x) \) is: \[ (-\infty, 1] \cup (2, 6] \] Thus, the domain of the function \( t \) is: \[ \boxed{(-\infty, 1] \cup (2, 6]} \]