\( g(x)=2 x+1 \quad x>1 \) \( h(x)=-x^{2}+1 \quad-1

Ask by Adkins Gross. in South Africa

Feb 26,2025

To find the range of the piecewise function defined by \( g(x) \) and \( h(x) \), we will analyze each function separately and then combine the results. 1. **Analyzing \( g(x) = 2x + 1 \) for \( x > 1 \)**: - As \( x \) approaches \( 1 \) from the right, \( g(1) = 2(1) + 1 = 3 \). - As \( x \) increases without bound, \( g(x) \) also increases without bound. - Therefore, the range of \( g(x) \) is \( (3, \infty) \). 2. **Analyzing \( h(x) = -x^2 + 1 \) for \( -1 < x < 1 \)**: - This is a downward-opening parabola with its vertex at \( (0, 1) \). - The maximum value occurs at \( x = 0 \), where \( h(0) = -0^2 + 1 = 1 \). - As \( x \) approaches \( -1 \) or \( 1 \), \( h(x) \) approaches \( -1 \) (since \( h(-1) = -(-1)^2 + 1 = 0 \) and \( h(1) = -(1)^2 + 1 = 0 \)). - Therefore, the range of \( h(x) \) is \( (-1, 1) \). 3. **Combining the ranges**: - The range of \( g(x) \) is \( (3, \infty) \). - The range of \( h(x) \) is \( (-1, 1) \). - Since there is no overlap between these two ranges, the overall range of the piecewise function is the union of the two ranges. Thus, the final range of the piecewise function is: \[ \text{Range} = (-1, 1) \cup (3, \infty) \]