QuestionAugust 25, 2025

2. Given the function f(x) = f(x)= ) sqrt (x+1)&if0leqslant xleqslant 3 2x^2-5&if3lt xleqslant 8 compute: f(2)= __ f(3)= __ f(4)=underline ( )

2. Given the function f(x) = f(x)= ) sqrt (x+1)&if0leqslant xleqslant 3 2x^2-5&if3lt xleqslant 8 compute: f(2)= __ f(3)= __ f(4)=underline ( )
2. Given the function f(x) = f(x)= ) sqrt (x+1)&if0leqslant xleqslant 3 2x^2-5&if3lt xleqslant 8 compute: f(2)= __ f(3)= __ f(4)=underline ( )

Solution
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Answer

f(2) = \sqrt{3} ### f(3) = 2 ### f(4) = 27 Explanation 1. Evaluate f(2) Since 0 \leq 2 \leq 3, use f(x) = \sqrt{x+1}. Calculate f(2) = \sqrt{2+1} = \sqrt{3}. 2. Evaluate f(3) Since 0 \leq 3 \leq 3, use f(x) = \sqrt{x+1}. Calculate f(3) = \sqrt{3+1} = \sqrt{4} = 2. 3. Evaluate f(4) Since 3 < 4 \leq 8, use f(x) = 2x^2 - 5. Calculate f(4) = 2(4)^2 - 5 = 32 - 5 = 27.

Explanation

1. Evaluate $f(2)$<br /> Since $0 \leq 2 \leq 3$, use $f(x) = \sqrt{x+1}$. Calculate $f(2) = \sqrt{2+1} = \sqrt{3}$.<br /><br />2. Evaluate $f(3)$<br /> Since $0 \leq 3 \leq 3$, use $f(x) = \sqrt{x+1}$. Calculate $f(3) = \sqrt{3+1} = \sqrt{4} = 2$.<br /><br />3. Evaluate $f(4)$<br /> Since $3 < 4 \leq 8$, use $f(x) = 2x^2 - 5$. Calculate $f(4) = 2(4)^2 - 5 = 32 - 5 = 27$.
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