Show that $f(x)=\dfrac{1}{x^2+1}$ is Lipschitz on interval $[-2,2]$, $L=$?
My work:
$$f'(x)=\frac{-2x}{(x^2+1)^2}$$
$$|f'(x)|=\left| \frac{-2x}{(x^2+1)^2} \right|=\frac{2|x|}{(x^2+1)^2}$$
Is $L=\dfrac{1}{2}$? Can I say that also $L$ can be any number bigger than $\dfrac{1}{2}$?
Ex: In this case $L=2$
Because the definition says that $|f(x)-f(y)| \le L| x - y |$ or $L$ needs to be the smallest possible? (Sorry for bad English)
Since $ f $ is even, the derivative on the positive and negative parts will only differ by a sign, as you stated. So you can just consider the maximum of $ f' $ on $ [0, 2] $ and take the absolute value.
$$ f' = \frac{-2x}{(x^2+1)^2} $$
$$ f'' = \frac{6x^2 - 2}{(x^2+1)^3} $$
Find where this equals 0 on [0, 2] to get the point where the absolute value of the derivative is at a maximum. $ 6x^2 - 2 = 0 $ when $ x = \frac{1}{\sqrt{3}} $ (the positive solution, though it doesn't really matter if you take the other solution because you will take the absolute value).
Using this, the maximum of $ | f' | $ is $ \frac{3 \sqrt{3}}{8} $.
Any secant line on this domain will have absolute value of slope less than or equal to this value, so this is the minimum Lipschitz constant for this function. $ L = \frac{1}{2} $ doesn't work because this maximum derivative is greater than $ \frac{1}{2} $. Any value of $ L $ which is greater than $ \frac{3 \sqrt{3}}{8} $ will still satisfy the inequality, but $ L = \frac{3 \sqrt{3}}{8} $ is the best possible bound.