QuestionAugust 26, 2025

3. Let g be a function defined by g(x) where g(x)= ) (-1)/(2)x^2+3,&ifxlt 1 2x^2+1,&ifxgeqslant 1 Find g(-1),g(0),g(1),g(2)

3. Let g be a function defined by g(x) where g(x)= ) (-1)/(2)x^2+3,&ifxlt 1 2x^2+1,&ifxgeqslant 1 Find g(-1),g(0),g(1),g(2)
3. Let g be a function defined by g(x) where g(x)= ) (-1)/(2)x^2+3,&ifxlt 1 2x^2+1,&ifxgeqslant 1 Find g(-1),g(0),g(1),g(2)

Solution
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Answer

g(-1) = \frac{5}{2}, g(0) = 3, g(1) = 3, g(2) = 9 Explanation 1. Evaluate g(-1) Since -1 < 1, use g(x) = -\frac{1}{2}x^2 + 3. Calculate g(-1) = -\frac{1}{2}(-1)^2 + 3 = -\frac{1}{2} + 3 = \frac{5}{2}. 2. Evaluate g(0) Since 0 < 1, use g(x) = -\frac{1}{2}x^2 + 3. Calculate g(0) = -\frac{1}{2}(0)^2 + 3 = 3. 3. Evaluate g(1) Since 1 \geq 1, use g(x) = 2x^2 + 1. Calculate g(1) = 2(1)^2 + 1 = 2 + 1 = 3. 4. Evaluate g(2) Since 2 \geq 1, use g(x) = 2x^2 + 1. Calculate g(2) = 2(2)^2 + 1 = 8 + 1 = 9.

Explanation

1. Evaluate $g(-1)$<br /> Since $-1 < 1$, use $g(x) = -\frac{1}{2}x^2 + 3$. Calculate $g(-1) = -\frac{1}{2}(-1)^2 + 3 = -\frac{1}{2} + 3 = \frac{5}{2}$.<br /><br />2. Evaluate $g(0)$<br /> Since $0 < 1$, use $g(x) = -\frac{1}{2}x^2 + 3$. Calculate $g(0) = -\frac{1}{2}(0)^2 + 3 = 3$.<br /><br />3. Evaluate $g(1)$<br /> Since $1 \geq 1$, use $g(x) = 2x^2 + 1$. Calculate $g(1) = 2(1)^2 + 1 = 2 + 1 = 3$.<br /><br />4. Evaluate $g(2)$<br /> Since $2 \geq 1$, use $g(x) = 2x^2 + 1$. Calculate $g(2) = 2(2)^2 + 1 = 8 + 1 = 9$.
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