I have a question about this following question:
Let $a>0$. Show that the maximum value of $f(x):=\frac{1}{1+|x|}+\frac{1}{1+|x-a|}$ is $\frac{2+a}{1+a}$
I am wondering if I am headed in the right direction with the following process. I first adress the issue of the absolute values by rewriting the function:
$f(x)=:\frac{1}{1+\sqrt{x^{2}}} +\frac{1}{1+\sqrt{(x-a)^{2}}}$
I then take the derivative of $f(x)$, which yields:
f'(x)=\frac{a-x}{(1+\sqrt{(a-x)^{2}})^{2}\sqrt{(a-x)^{2}}}+\frac{x}{\sqrt{x^{2}}(1+\sqrt{x^{2}})^{2}}
The derivative I've taken from Wolfram Alpha for the moment. I am a bit confused about my next step, though I have an idea. Would I input $\frac{2+a}{1+a}$ into the first derivative, which should give me back zero? This would show that there is a maximum or minimum at that point. The next step being to check the value of the second derivative to show that it is indeed a maximum? I suppose this still wont show that $\frac{2+a}{1+a}$ is the global maximum.