Question

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}`?

Answer 1

Yes it is.

Explanation:

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.