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

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 .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 ,∞)

```
8 years ago
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