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

Mar 29, 2017

$f \left(x\right) = \left\mid x - 1 \right\mid \sqrt{x}$

is continuous in $\left[0 , + \infty\right)$ and differentiable in $\left[0 , 1\right) \cup \left(1 , + \infty\right)$

#### Explanation:

We have that:

$x < 1 \implies x - 1 < 0 \implies \left\mid x - 1 \right\mid = - \left(x - 1\right)$

$x > 1 \implies x - 1 > 0 \implies \left\mid x - 1 \right\mid = \left(x - 1\right)$

so:

• For $x \in \left[0 , 1\right)$, $f \left(x\right) = \left(- x + 1\right) \sqrt{x}$
• For $x = 1$, $f \left(x\right) = 0$
• For $x \in \left(1 , + \infty\right)$, $f \left(x\right) = \left(x - 1\right) \sqrt{x}$

We can deduce that the function is continuous in $\left[0 , 1\right) \cup \left(1 , + \infty\right)$ since it is the composition of continuous functions.

For $x = 1$ we have:

${\lim}_{x \to {1}^{-}} f \left(x\right) = {\lim}_{x \to {1}^{-}} \left(- x + 1\right) \sqrt{x} = 0$

${\lim}_{x \to {1}^{+}} f \left(x\right) = {\lim}_{x \to {1}^{+}} \left(x - 1\right) \sqrt{x} = 0$

which implies:

${\lim}_{x \to 1} f \left(x\right) = 0 = f \left(1\right)$

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

Differentiating $f \left(x\right)$ we have:

• For $x \in \left[0 , 1\right)$, $f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\left(- x + 1\right) \sqrt{x}\right) = - \sqrt{x} - \frac{x + 1}{2 \sqrt{x}} = \frac{1 - 3 x}{2 \sqrt{x}}$
• For $x \in \left(1 , + \infty\right)$, $f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\left(x - 1\right) \sqrt{x}\right) = \sqrt{x} + \frac{x - 1}{2 \sqrt{x}} = \frac{3 x - 1}{2 \sqrt{x}}$

so for $x = 1$:

${\lim}_{x \to {1}^{-}} f ' \left(x\right) = {\lim}_{x \to {1}^{-}} \frac{1 - 3 x}{2 \sqrt{x}} = - 1$

${\lim}_{x \to {1}^{+}} f ' \left(x\right) = {\lim}_{x \to {1}^{+}} \frac{3 x - 1}{2 \sqrt{x}} = 1$

As the derivative is not continuous in $x = 1$ the function $f \left(x\right)$ is not differentiable.