Frank Guyton

2022-01-03

Prove that $|\mathrm{log}(1+{x}^{2})-\mathrm{log}(1+{y}^{2})|\le |x-y|$

Annie Gonzalez

Beginner2022-01-04Added 41 answers

The derivative of $f\left(x\right)=\mathrm{log}(1+{x}^{2})$ is

$f}^{\prime}\left(x\right)=\frac{2x}{1+{x}^{2}$

and

$\left|\frac{2x}{1+{x}^{2}}\right|\le 1$

because

$2\left|x\right|\le 1+{\left|x\right|}^{2}$

since

${(1-\left|x\right|)}^{2}\ge 0$

Thus, by the mean value theorem,

$\frac{\mathrm{log}(1+{x}^{2})-\mathrm{log}(1+{y}^{2})}{x-y}={f}^{\prime}\left(z\right)$

for$z$ between $x$ and $y$ (assuming $x\ne y$ or the inequality is obvious). Then

$\left|\frac{\mathrm{log}(1+{x}^{2})-\mathrm{log}(1+{y}^{2})}{x-y}\right|=|{f}^{\prime}\left(z\right)\le 1|$

and

because

since

Thus, by the mean value theorem,

for

Travis Hicks

Beginner2022-01-05Added 29 answers

HINT: using the mean value theorem for the function

$f\left(t\right)=\mathrm{ln}(1+{t}^{2})$

we get

$f\left(x\right)-f\left(y\right)=\frac{2t}{1+{t}^{2}}(x-y)$

and we have

$\left|\frac{2t}{1+{t}^{2}}\right|\le 1$

we get

and we have

Vasquez

Expert2022-01-11Added 669 answers

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