We have #ainRR#* and # f:RR#\#{1}->RR, f(x)=sqrt(x^2+a^2)/(x-1)#.Is this function monotone on #(0,oo)#\#{1}#?

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Answer 1 · 2018-07-13T14:40:17

Yes it is.

We will compute the first derivative, using the Quotient and the chain rule:

#f'(x)=(1/2*(x^2+a^2)^(-1/2)*2x*(x-1)-sqrt(x^2+a^2))/(x-1)^2#

Multiplying denominator and numerator by #sqrt(x^2+a^2)# we get

#f'(x)=(1/2(x-1)*2x-x^2-a^2)/(sqrt(a^2+x^2)*(x-1)^2)#

simplifying we get

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

so #f'(x)# is negative on the given interval, so #f(x)# is monotonously decreasing.