Question 7 Let \( f(x)=2 x+1, g(x)=-x^{2}+1 \), and \( h(x)=\frac{1}{x+1} \). The function \( t \) is defined as \( f(x)=\left\{\begin{array}{ll}h(x) & \text { if } x \in(2,6] \\ g(x) & \text { if } x \leq 1\end{array}\right. \) The domain of function \( t \) is: (A) \( \mathrm{Dt}=(-\infty, 1] \cup(2,6) \) (B) \( \mathrm{Dt}=(-\infty, 6] \) (C) None of these (D) \( \mathrm{Dt}=(-\infty, 1) \cup[, 6] \) (E) \( \mathrm{Dt}=(-\infty, 1] \cup[2,6] \)

To determine the domain of the function \( t \), we need to analyze the conditions under which \( t \) is defined based on the piecewise definition provided. The function \( t \) is defined as follows: \[ t(x) = \begin{cases} h(x) & \text{if } x \in (2, 6] \\ g(x) & \text{if } x \leq 1 \end{cases} \] ### Step 1: Analyze the conditions for \( h(x) \) The function \( h(x) = \frac{1}{x+1} \) is defined for all \( x \) except where the denominator is zero. This occurs when: \[ x + 1 = 0 \implies x = -1 \] Thus, \( h(x) \) is defined for \( x \in (-\infty, -1) \cup (-1, \infty) \). Since \( t(x) = h(x) \) for \( x \in (2, 6] \), we need to check if \( -1 \) falls within this interval. It does not, so \( h(x) \) is valid for all \( x \) in \( (2, 6] \). ### Step 2: Analyze the conditions for \( g(x) \) The function \( g(x) = -x^2 + 1 \) is a polynomial and is defined for all real numbers. Therefore, \( g(x) \) is valid for \( x \leq 1 \). ### Step 3: Combine the intervals Now we combine the intervals where \( t(x) \) is defined: 1. From \( g(x) \): \( x \leq 1 \) gives us the interval \( (-\infty, 1] \). 2. From \( h(x) \): \( x \in (2, 6] \) gives us the interval \( (2, 6] \). ### Step 4: Determine the overall domain The overall domain of \( t \) is the union of the two intervals: \[ \text{Domain of } t = (-\infty, 1] \cup (2, 6] \] ### Conclusion Now, we can match this result with the provided options: (A) \( \mathrm{Dt}=(-\infty, 1] \cup(2,6) \) (B) \( \mathrm{Dt}=(-\infty, 6] \) (C) None of these (D) \( \mathrm{Dt}=(-\infty, 1) \cup[, 6] \) (E) \( \mathrm{Dt}=(-\infty, 1] \cup[2,6] \) The correct answer is: **(A) \( \mathrm{Dt}=(-\infty, 1] \cup(2,6) \)**.