Find the range of the function f(x) = (1+ x^2)/x^2 ?

1 Answer
Jim S
May 10, 2018

#f(A) = (1,+oo)#

Explanation:

#f(x)=(x^2+1)/x^2# , #A=(-oo,0)uu(0,+oo)#

#f'(x)=((x^2+1)'x^2-(x^2)'(x^2+1))/x^4=#

#(2x^3-2x^3-2x)/x^4=#

#-2/x^3#

For #x>0# we have #f'(x)<0# so #f# is strictly decreasing in #(0,+oo)#

For #x<0# we have #f'(x)>0# so #f# is strictly increasing in #(-oo,0)#

#A_1=(-oo,0)#, #A_2=(0,+oo)#

#lim_(xrarr0^(-))f(x)=lim_(xrarr0^(-))(x^2+1)/x^2=+oo#
#lim_(xrarr0^(+))f(x)=lim_(xrarr0^(+))(x^2+1)/x^2=+oo#
#lim_(xrarr-oo)f(x)=lim_(xrarr-oo)(x^2+1)/x^2=lim_(xrarr-oo)x^2/x^2=1#
#lim_(xrarr+oo)f(x)=lim_(xrarr+oo)(x^2+1)/x^2=1#

#f(A_1)=f(((-oo,0)))=(lim_(xrarr-oo)f(x),lim_(xrarr0^(-))f(x))=#

#(1,+oo)#

#f(A_2)=f(((0,+oo)))=(lim_(xrarr+oo)f(x),lim_(xrarr0^+)f(x))=(1,+oo)#

Range #=f(A)=f(A_1)uuf(A_2)=(1,+oo)#

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