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

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Answer 1 · 2018-05-13T13:34:24

The domain is #x in RR#. The range is #y in[-0.04, 1.24]#

The denominator is always #>0# whatever #x in RR#

The domain is #x in RR#

To find the domain, proceed as follows

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

Rearranging,

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

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

This is a quadratic equation in #x# and in order for this equation to have solutions, the discriminant #>=0#

#Delta=(-1)^2-4*y(5y-6)=1-20y^2+24y#

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

The solutions to this inequality is

#y in [(-24+sqrt(24^2+4*20))/(-40),(-24-sqrt(24^2+4*20))/(-40)]#

#y in[-0.04, 1.24]#

The range is #y in[-0.04, 1.24]#

graph{(x+6)/(x^2+5) [-10.47, 3.574, -3.903, 3.117]}