Question

已知f(x)=根号下1+x2 ,a≠b,求证∣f(a)-f(b)∣＜∣a-b∣

Answer 1

f(x)=√(1+x^2)

|f(a)-f(b)|=|√(1+a^2))-√(1+b^2)|

=|((1+a^2))-(1+b^2))/(√(1+a^2))+√(1+b^2))|

=|(a+b)(a-b)/(√(1+a^2))+√(1+b^2))|

=|a-b|*|(a+b)|/(√(1+a^2))+√(1+b^2))

≤|a-b|*(|a|+|b|)/(√(1+a^2))+√(1+b^2))

≤|a-b|*(|a|+|b|)/(|a|+|b|)

=|a-b|