let a0. Show that the maximum value of h(x) is (2+a)/(1+a)

h(x) = 1/(1+abs(x)) + 1/(1+abs(x-a))

Question

let a>0.  Show that the maximum value of h(x) is (2+a)/(1+a)

h(x) = 1/(1+abs(x)) + 1/(1+abs(x-a))

anonymous · Answer

x>a$$h(x)=\frac{1}{1+x}+\frac{1}{1+x-a}$$0<x<a$$h(x)=\frac{1}{1+x}+\frac{1}{1+a-x}$$x<0$$h(x)=\frac{1}{1-x}+\frac{1}{1-x-a}$$

anonymous · Answer

and i have a question @Stamished are u allowed to use calculus?

anonymous · Answer

yes, definitely.  I'd actually prefer if I could get an answer with Calculus (:

anonymous · Answer

thanks so much for helping

anonymous · Answer

np  :)

anonymous · Answer

and dont forget to check bounds... x=0 and x=a

anonymous · Answer

hmmm..., I'm still confused on what I should do next...

anonymous · Answer

differentiate on each interval and find critical points...

anonymous · Answer

someone feel free to offer a better approach.

anonymous · Answer

kay, I'm going to try it now

anonymous · Answer

for example  $x\ge a$ $$h'(x)=-\frac{1}{(1+x)^2}-\frac{1}{(1+x-a)^2}<0$$decreasing for all x>a so the maximum of this interval take places at x=a

u do the rest :)

anonymous · Answer

okay cool, so I'm doing it right so far. Thanks!! :D