The range of the function

ƒ(x)=8^x+4^x+8^-x+4^-x+5 is

hi,

 you can see that ,

                   ƒ(x)= ƒ(-x)

it indicates that our graph is symmetrical  on both sides ofy- axis, so analysis of function on any of the side will be enough,  we will take care of only +x axis.....

  ƒ(x)=8^x+4^x+8^-x+4^-x+5   ,

on diff. w.r.t x

d/df  ƒ(x)  =  ln 8 .8^x + ln 4 .4^{x  _{- ln8}} 8^{-x  _{- ln4}} 4^-x 

                = ln8 ( 8^{x _-} 8^{-x )   _{+  ln 4 (}}  .4^{x  _-} 4^-x )

^{we know that if  A > 1 than A -1/A   will always be >0}

^{_{so ,}} ( 8^{x _-} 8^{-x )    _{&  (}} 4^{x  _-} 4^{-x )  _{will always be >0}}

^_so,  ln8 ( 8^{x _-} 8^{-x )   _{+  ln 4 (}}  .4^{x  _-} 4^{-x )    >0    for    x  >0       _{( remind that we are taking care of only  +x axis )}}

_{^{it shows that df/dx is always >o  for x >o}}

_{^{means the min value would be at x =0}} 

                              _{^{min. f   = f (o) =  9}}

_{^{& its obvious that the max. value of F(X) is  infinity}}

  _{^{range of f(x) =  [ 9 ,∞)}}