Question

How to solve this? Determine the continuity and derivability domain for `f(x)` #f:[0,oo)->RR,f(x)=|x-1|sqrtx #

Answer 1

`f(x) = abs(x-1)sqrtx`

is continuous in `[0,+oo)` and differentiable in `[0,1) uu (1,+oo)`

Explanation:

We have that:

`x<1 => x-1 < 0 => abs(x-1) = -(x-1)`

`x > 1 => x-1 > 0 => abs(x-1) = (x-1)`

so:

• For `x in [0,1)`, `f(x) = (-x+1)sqrt(x)`
• For `x = 1`, `f(x) = 0`
• For `x in (1,+oo)`, `f(x) = (x-1)sqrt(x)`

We can deduce that the function is continuous in `[0,1) uu (1,+oo)` since it is the composition of continuous functions.

For `x=1` we have:

`lim_(x->1^-) f(x) = lim_(x->1^-) (-x+1)sqrt(x) = 0`

`lim_(x->1^+) f(x) = lim_(x->1^+) (x-1)sqrt(x) = 0`

which implies:

`lim_(x->1) f(x) = 0 = f(1)`

so the function is continuous also in `x=1`.

Differentiating `f(x)` we have:

• For `x in [0,1)`, `f'(x) = d/dx((-x+1)sqrt(x)) = -sqrtx -(x+1)/(2sqrt(x)) = (1-3x)/(2sqrtx)`
• For `x in (1,+oo)`, `f'(x) = d/dx((x-1)sqrt(x)) = sqrt(x) + (x-1)/(2sqrtx) =(3x-1)/(2sqrtx)`

so for `x=1`:

`lim_(x->1^-) f'(x) = lim_(x->1^-) (1-3x)/(2sqrtx) =- 1`

`lim_(x->1^+) f'(x) = lim_(x->1^+) (3x-1)/(2sqrtx) = 1`

As the derivative is not continuous in `x=1` the function `f(x)` is not differentiable.