What is the domain and range of #(x+5)/(x^2+36)#?

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Answer 1 · 2017-11-15T13:13:18

The domain is #x in RR#.
The range is #y in [-0.04,0.18]#

The denominator is #>0#

#AA x in RR#, #x^2+36>0#

Therefore,

The domain is #x in RR#

Let,

#y=(x+5)/(x^2+36)#

Simplifying and rearranging

#y(x^2+36)=x+5#

#yx^2-x+36y-5=0#

This is a quadratic equation in #x^2#

In order for this equation to have solutions, the discriminant #Delta>=0#

So,

#Delta=b^2-4ac=(-1)^2-4(y)(36y-5)>=0#

#1-144y^2+20y>=0#

#144y^2-20y-1<=0#

#y=(20+-sqrt(400+4*144))/(288)#

#y_1=(20+31.24)/188=0.18#

#y_2=(20-31.24)/288=-0.04#

Therefore,

The range is #y in [-0.04,0.18]#

graph{(x+5)/(x^2+36) [-8.89, 8.884, -4.44, 4.44]}