Question #256751
Find the domain and range of the following rational function. Use qny notation. 1) f(x) = 2 / x+1 2) f(x) = 3x / x+3 3) f(x) = 3-x / x-7 4) f(x) = 2+x / x 5) f(x) = (x+1) / x^2-1
1
Expert's answer
2021-10-26T17:29:49-0400

1) f(x) = 2x+1\frac{2}{x+1}


Here x+1 \not = 0 // otherwise function will not be defined

So x \not = - 1


Domain (,1)-\infty,-1) \bigcup (1,)(-1 , \infty)


Range (,0)(0,)(-\infty ,0) \bigcup (0,\infty)



2) f(x) = 3xx+3\frac{3x}{x+3}


Here x+3 \not = 0 // otherwise function will not be defined

So x \not = -3


Domain (,3)(3,)(-\infty ,-3) \bigcup (-3 ,\infty)


f(x) = 3x+99x+3\frac{3x +9 -9}{x+3} = 3(x+3)9x+3\frac{3(x+3) - 9}{x+3} = 3 - 9x+3\frac{9}{x+3} ----------------------- (1)


From equation (1) we can understand the value will never reach 3.


So Range (,3)(3,)(-\infty ,3) \bigcup (3, \infty)


3) f(x) = 3xx7\frac{3-x}{x-7}


Here x-7 \not = 0 // otherwise function will not be defined


So x \not = 7


Domain (,7)(7,)(-\infty ,7) \bigcup (7,\infty)


f(x) = (x3)x7\frac{-(x-3)}{x-7} = (x34+4)(x7)\frac{-(x-3-4+4)}{(x-7)} =(x7+4)(x7)\frac{-(x-7+4)}{(x-7)} = -1 +4(x7)\frac{-4}{(x-7)}


Range (,1)(1,)(-\infty ,-1) \bigcup (-1,\infty)



4) f(x) = 2+xx\frac{2+x}{x}


Here x \not = 0


Domain (,0)(0,)(-\infty ,0) \bigcup (0,\infty)


f(x) = 2+xx\frac{2+x}{x} = 2x\frac{2}{x} + 1


Range (,1)(1,)(-\infty,1) \bigcup (1,\infty)


5) f(x) = x+1x21\frac{x+1}{x^2 -1}


Here x2 - 1 \not = 0

So x 1\not = 1 and x 1\not = -1


Domain (,1)(1,1)(1,)(-\infty,-1) \bigcup (-1,1) \bigcup (1,\infty)


Range (,0)(0,)(-\infty ,0) \bigcup (0,\infty)





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