Draw the graph of `f(x)= [sqrtx], x in [0, 16)`, where `[*]` denotes the greatest ineger function.

Explanation

(i) `f(x)=[sqrt(x)], x ge 0.`
`[sqrt(x)]=0 " if " sqrt(x) in [0,1) " or " x in [0,1)`
`[sqrt(x)]=1 " if " sqrt(x) in [1,2) " or " x in [1,4)`
`[sqrt(x)]=2 " if " sqrt(x) in [2,3) " or " x in [4,9)`
and so on.
` :. f(x) ={(0"," x in[0,1)),(1"," x in[1,4)),(2"," x in[4,9)),(3"," x in[9,16)),(...),(...):}`
(ii) `f(x)=[tan^(-1)x]`
We know that `tan^(-1)x in(-(pi)/(2),(pi)/(2))`
` :. " possible values of " [tan^(-1) x]" are " -2,-1,0,1.`
If `[tan^(-1)x]= -2," then " tan^(-1)x in (-(pi)/(2),-1) " or " x in (-oo,-tan1) `
If `[tan^(-1)x]= -1," then " tan^(-1)x in [-1,0) " or " x in [-tan1,0) `
If `[tan^(-1)x]= 0," then " tan^(-1)x in [0,1) " or " x in [0,tan 1) `
If `[tan^(-1)x]= 1," then " tan^(-1)x in [1,(pi)/(2)) " or " x in [tan 1, oo) `
(iii) `f(x) =[log_(e)x]`
We know that `log_(e) in (-oo,oo)`.
` :. [log_(e)x] in Z," i.e., " [log_(e)x]` takes all integral values.
If `[log_(e)x] =n,n in Z " then " log_(e)x in [n,n+1) " or " x in [e^(n),e^(n+1))`.
Therefore, `[log_(e)x]={(...),(...),(-1"," x in[e^(-1),1)),(0", " x in[1,e)),(1", " x in[e,e^(2))),(2", " x in[e^(2),e^(3))),(3", " x in[e^(3),e^(4))),(...),(...):}`