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The range of the function ƒ(x)=8x+4x+8-x+4-x+5 is
The range of the function
ƒ(x)=8x+4x+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)=8x+4x+8-x+4-x+5 , on diff. w.r.t x d/df ƒ(x) = ln 8 .8x + ln 4 .4x - ln8 8-x - ln4 4-x = ln8 ( 8x - 8-x ) + ln 4 ( .4x - 4-x ) we know that if A > 1 than A -1/A will always be >0 so , ( 8x - 8-x ) & ( 4x - 4-x ) will always be >0 so, ln8 ( 8x - 8-x ) + ln 4 ( .4x - 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 ,∞)
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)=8x+4x+8-x+4-x+5 ,
on diff. w.r.t x
d/df ƒ(x) = ln 8 .8x + ln 4 .4x - ln8 8-x - ln4 4-x
= ln8 ( 8x - 8-x ) + ln 4 ( .4x - 4-x )
we know that if A > 1 than A -1/A will always be >0
so , ( 8x - 8-x ) & ( 4x - 4-x ) will always be >0
so, ln8 ( 8x - 8-x ) + ln 4 ( .4x - 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 ,∞)
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