Question

Let $f(x) = x^{3/2} - \sqrt{x^3 + x^2}$ then

• A. f(x) is non differentiable at x = 2
• B. f(x) is discontinuous at x = 5
• C. f(x) is discontinuous and non differentiable at x = 1
• D. f(x) is continuous and differentiable at x = 0

Answer 1

Answer: (D)

f(x) is continuous and differentiable at x = 0

Answer 2

We have:

$f(x)= x^{3/2} - \sqrt{x^3+x^2} = x^{3/2} - \sqrt{x^2(x+1)} = x^{3/2} - x\sqrt{x+1} \quad (x\ge0)$

Checking each option:

1. At $x=2$:
   $f(x)$ is made up of elementary functions (powers and roots) and is differentiable since its derivative exists: $f'(x)= \frac{3}{2}x^{1/2} - \left(\sqrt{x+1} + \frac{x}{2\sqrt{x+1}}\right)$ At $x=2$, this expression is finite. So non-differentiability at 2 is false.

2. At $x=5$:
   All functions involved are continuous for $x\ge0$. So discontinuity at 5 is false.

3. At $x=1$:
   $f(1)= 1^{3/2} - 1\sqrt{1+1} = 1 - \sqrt{2}$ is continuous. Its derivative at $x=1$: $f'(1)= \frac{3}{2} - \sqrt{2} - \frac{1}{2\sqrt{2}}$ exists and is finite. Hence, discontinuity and non-differentiability at 1 is false.

4. At $x=0$:
   Since $f(0)= 0^{3/2} - 0\sqrt{0+1}=0$, $f(x)$ is continuous at $x=0$. For differentiability, using the right-hand derivative (since the domain is $x\ge 0$): $f'(0)= \lim_{h\to 0^+} \frac{f(h)-f(0)}{h} = \lim_{h\to 0^+} \left(h^{1/2} - \sqrt{h+1}\right)= 0-1=-1$. Thus, $f(x)$ is differentiable at $x=0$.

Conclusion: Only option 4 is correct.