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Exercise 1.3 · Q3

Q.Write the values of ff at −3, 5, 2, −1, 0-3,\ 5,\ 2,\ -1,\ 0 if
[!FORMULA] f(x)={x2+x−5if x∈(−∞,0)x2+3x−2if x∈(3,∞)x2if x∈(0,2)x2−3otherwisef(x)=\begin{cases}x^2+x-5 & \text{if } x\in(-\infty,0)\\ x^2+3x-2 & \text{if } x\in(3,\infty)\\ x^2 & \text{if } x\in(0,2)\\ x^2-3 & \text{otherwise}\end{cases}

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✓ Free question

Step 1. x=−3x=-3: since −3∈(−∞,0)-3\in(-\infty,0), use f(x)=x2+x−5f(x)=x^2+x-5: f(−3)=9−3−5=1f(-3)=9-3-5=1.

Step 2. x=5x=5: since 5∈(3,∞)5\in(3,\infty), use f(x)=x2+3x−2f(x)=x^2+3x-2: f(5)=25+15−2=38f(5)=25+15-2=38.

Step 3. x=2x=2: the piece x∈(0,2)x\in(0,2) is an OPEN interval, so x=2x=2 is excluded from it; 22 is also not in (−∞,0)(-\infty,0) or (3,∞)(3,\infty). So x=2x=2 falls to "otherwise": f(x)=x2−3⇒f(2)=4−3=1f(x)=x^2-3\Rightarrow f(2)=4-3=1.

Step 4. x=−1x=-1: since −1∈(−∞,0)-1\in(-\infty,0), use f(x)=x2+x−5f(x)=x^2+x-5: f(−1)=1−1−5=−5f(-1)=1-1-5=-5.

Step 5. x=0x=0: (0,2)(0,2) is open so it excludes 00; (−∞,0)(-\infty,0) also excludes 00. So x=0x=0 falls to "otherwise": f(0)=0−3=−3f(0)=0-3=-3.

✓Final answer

f(−3)=1,f(5)=38,f(2)=1,f(−1)=−5,f(0)=−3f(-3)=1,\quad f(5)=38,\quad f(2)=1,\quad f(-1)=-5,\quad f(0)=-3.

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