How do you find the domain and range of #f(x)=(x+7)/(x^2-49)#?

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Answer 1 · 2017-10-07T16:28:38

The domain is #x in RR-{7}#
The range is #f(x) in (-oo,0^-)uu(0^+,+oo)#

Let's simplify the function

#f(x)=(x+7)/(x^2-49)=cancel(x+7)/(cancel(x+7)(x-7))#

#f(x)=1/(x-7)#

The denominator is #!=0#

Therefore,

#x-7!=0#, #=>#,#x!=7#

The domain is #I=RR-{7}#

The function is bijective over the domain #I#

#f(-oo)=1/(-oo-7)=0^-#

#f(7^-)=1/(7^(-) -7)=-oo#

#f(7^+)=1/(7^(+) -7)=+oo#

#f(+oo)=1/(+oo-7)=0^+#

Therefore,

The range is #f(x) in (-oo,0^-)uu(0^+,+oo)#

graph{(1/(x-7)) [-13.34, 27.2, -10.06, 10.22]}