Question

The set of all points of differentiability of the function $f\left(x\right)=\frac{\sqrt{x+1}-1}{\sqrt{x}}$ for $x\ne 0$ and $f\left(0\right)=0$ is

• A. $(-\infty ,\infty )$
• B. $[0,\infty )$
• C. $(0,\infty )$
• D. $(-\infty ,\infty )\sim \left\{0\right\}$

Answer 1

Answer: (C)

$(0,\infty )$

For the function to be differentiable the function should exists at the first case in the real numbers domain.

The function is:

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

For the function to exists from $\sqrt{x+1}$ we have the relation $x\ge -1$ and the from the number $\sqrt{x}$ we have $x>0$. The intersection of both the regions is $x>0$.

Now differentiating it once to check at which points the function is not differentiable.

$f^{\prime }\left(x\right)=\frac{\frac{\sqrt{x+1}-1}{2\sqrt{x}}-\frac{\sqrt{x}}{2\sqrt{x+1}}}{x}$

$f^{\prime }\left(x\right)=\frac{1}{2}\frac{x+1-\sqrt{x+1}-x}{x^{3/2}\sqrt{x+1}}$

$f^{\prime }\left(x\right)=\frac{1-\sqrt{x+1}}{2x^{3/2}\sqrt{x+1}}$

For the derivative of the function to exists from $\sqrt{x+1}$ we have the relation $x\ge -1$ and the from the number $\sqrt{x}$ we have $x>0$. The intersection of both the regions is $x>0$. Hence the region is $(0,\infty )$. ....Answer