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

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Answer 1 · 2017-03-29T14:35:49

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

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

We have that:

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

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

so:

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.

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