The range of the function &fnof;(x)=8 x +4 x +8 -x +4 -x +5 is

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Pratham Ashish · Answer

hi, you can see that , &fnof;(x)= &fnof;(-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..... &fnof;(x)=8 x +4 x +8 -x +4 -x +5 , on diff. w.r.t x d/df &fnof;(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 ,&infin;)

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The range of the function ƒ(x)=8 x +4 x +8 -x +4 -x +5 is

The range of the function

ƒ(x)=8^x+4^x+8^-x+4^-x+5 is

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