Question

Question #5571f

Answer 1

`R_f=(-oo,1]`

Explanation:

`f(x)=1-sqrtx` ,

• `x>=0`

`D_f=[0,+oo)`

`f` is continuous in `D_f`

For `x` `in` `D_f`,

`f'(x)=-1/(2sqrtx)` `<0` , if `x>0`

so `f` is strictly decreasing in `[0,+oo)`

`R_f`=`f(D_f)=f([0,+oo))=(lim_(xrarr+oo)f(x),f(0)]=(-oo,1]`

because

• `lim_(xrarr+oo)f(x)=lim_(xrarr+oo)(1-sqrtx)=-(+oo)=-oo`

• `lim_(xrarr+oo)x=+oo`
  and so
  `lim_(xrarr+oo)sqrtx=+oo`