Differential Calculus-1
Differential Calculus-1
Allen Star Batch
Grade 12
Question:
The function $f(x) = \sqrt{1 - \sqrt{1 - x^2}}$
has its domain $-1 \leq x \leq 1$
has finite one sided derivatives at the point $x = 0$
is continuous and differentiable at $x = 0$
is continuous but not differentiable at $x = 0$
Step-by-Step Solution
Key Concept: For absolute value functions, check left and right derivatives separately; they must be equal for differentiability.
The function $f(x) = \frac{|x|}{\sqrt{1 + \sqrt{1 - x^2}}}$ has right derivative at $x = 0$ computed as $f'(0^+) = \lim_{h \to 0^+} \frac{h/(\sqrt{1+\sqrt{1-h^2}})}{h} = \frac{1}{\sqrt{2}}$. Similarly, the left derivative is $f'(0^-) = \lim_{h \to 0^-} \frac{h/(\sqrt{1+\sqrt{1-x^2}})}{-h} = -\frac{1}{\sqrt{2}}$. Since the left and right derivatives differ, the function is not differentiable at $x = 0$.
Correct Answer: 1,2,4