find the domain and the range of the following rational function.use any of the 3 notation
4.f(x)= 2+x/2x
5. f(x)= (x2+4x+3)/x²-9
Answers
Answer:
Expert's answer
1. f(x)=\dfrac{2}{x+1}f(x)=
x+1
2
x+1\not=0=>x\not=-1x+1
=0=>x
=−1
Domain:(-\infin, -1)\cup (-1, \infin)Domain:(−∞,−1)∪(−1,∞)
Range:(-\infin, 0)\cup (0, \infin)Range:(−∞,0)∪(0,∞)
2. f(x)=\dfrac{3x}{x+3}f(x)=
x+3
3x
x+3\not=0=>x\not=-3x+3
=0=>x
=−3
f(x)=\dfrac{3x}{x+3}=\dfrac{3x+9-9}{x+3}=3-\dfrac{9}{x+3}f(x)=
x+3
3x
=
x+3
3x+9−9
=3−
x+3
9
Domain:(-\infin, -3)\cup (-3, \infin)Domain:(−∞,−3)∪(−3,∞)
Range:(-\infin, 3)\cup (3, \infin)Range:(−∞,3)∪(3,∞)
3. f(x)=\dfrac{3-x}{x-7}f(x)=
x−7
3−x
x-7\not=0=>x\not=7x−7
=0=>x
=7
f(x)=\dfrac{3-x}{x-7}=\dfrac{3-(x-7)-7}{x-7}=-1-\dfrac{4}{x-7}f(x)=
x−7
3−x
=
x−7
3−(x−7)−7
=−1−
x−7
4
Domain:(-\infin, 7)\cup (7, \infin)Domain:(−∞,7)∪(7,∞)
Range:(-\infin, -1)\cup (-1, \infin)Range:(−∞,−1)∪(−1,∞)
4. f(x)=\dfrac{2+x}{x}f(x)=
x
2+x
x\not=0x
=0
f(x)=\dfrac{2+x}{x}=1+\dfrac{2}{x}f(x)=
x
2+x
=1+
x
2
Domain:(-\infin, 0)\cup (0, \infin)Domain:(−∞,0)∪(0,∞)
Range:(-\infin, 1)\cup (1, \infin)Range:(−∞,1)∪(1,∞)
5. f(x)=\dfrac{x+1}{x^2-1}f(x)=
x
2
−1
x+1
x^2-1\not=0=>x\not=-1, x\not=1x
2
−1
=0=>x
=−1,x
=1
f(x)=\dfrac{x+1}{x^2-1}=\dfrac{x+1}{(x+1)(x-1)}=-\dfrac{1}{x-1}, x\not=\pm1f(x)=
x
2
−1
x+1
=
(x+1)(x−1)
x+1
=−
x−1
1
,x
=±1
Domain:(-\infin,-1)\cup (-1,1)\cup (1, \infin)Domain:(−∞,−1)∪(−1,1)∪(1,∞)
Range:(-\infin, 0)\cup (0, \infin)Range:(−∞,0)∪(0,∞)
Answer:
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