How do you find the domain and range of #f(x)=(x+6)/(x^2+5) #?

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Answer 1 · 2018-06-26T06:31:27

The domain is #=RR#.
The range is #y in [-0.4,1.24 ] #

The function is

#f(x)=(x+6)/(x^2+5)#

#AA x in RR#, the denominator is #x^2+5>0#

The domain is #=RR#

To calculate the range, let

#y=(x+6)/(x^2+5)#

#=>#, #y(x^2+5)=x+6#

#=>#, #yx^2-x+5y-6=0#

This is a quadratic equation in #x^2# and in order to have solutions, the discriminant must be #>=0#

Therefore,

#Delta=(-1)^2-4(y)(5y-6)>=0#

#=>#, #1-20y^2+24y>= 0#

#=>#, #20y^2-24y-1<=0#

The solutions to this inequality is

#y in [(24-sqrt(24^2+4*20))/(2*20), (24+sqrt(24^2+4*20))/(2*20)]#

#y in [-0.4,1.24 ]#

The range is #y in [-0.4,1.24 ] #

graph{(x+6)/(x^2+5) [-13.7, 14.78, -7.27, 6.97]}