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ANKUR GUPTA Grade: 12
        

The range of the function


ƒ(x)=8x+4x+8-x+4-x+5  is

7 years ago

Answers : (1)

Pratham Ashish
17 Points
										

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 .4- ln8 8-x  - ln4 4-x 


                = ln8 ( 8x - 8-x )   +  ln 4 (  .4- 4-x )



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


so , ( 8x - 8-x )    &  ( 4- 4-x )  will always be >0


so,  ln8 ( 8x - 8-x )   +  ln 4 (  .4- 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 ,∞) 


7 years ago
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