Question

Consider the function $f : (0, 2) \to \mathbb{R}$ defined by $f(x) = \frac{x}{2} + \frac{2}{x}$ and the function $g(x)$ defined by $g(x) = \begin{cases} \min\\{f(t)\\}, & 0 < t \leq x \text{ and } 0 < x \leq 1 \\\ \frac{3}{2} + x, & 1 < x < 2 \end{cases}$ Then:

• A. g is continuous but not differentiable at x = 1
• B. g is not continuous for all x (0,2)
• C. g is neither continuous nor differentiable at x = 1
• D. g is continuous and differentiable for all x (0,2) 

Answer 1

Answer: (A)

g is continuous but not differentiable at x = 1

Answer 2

To determine the continuity and differentiability of $g$ at $x = 1$, we need to check the left-hand limit (LHL) and right-hand limit (RHL) as well as the derivative behavior at $x = 1$.

Continuity Check:

For $0 < x \leq 1$, $g(x) = \min\\{f(t)\\}$ where $f(t) = \frac{t}{2} + \frac{2}{t}$.

At $x = 1$, $f(1) = \frac{1}{2} + 2 = \frac{5}{2}$.

For $1 < x < 2$, $g(x) = \frac{3}{2} + x$.

- Left-hand limit as $x$ approaches $1$ from the left:
\lim_{x \to 1^-} g(x) = \min\left\\{\frac{5}{2}\right\\} = \frac{5}{2}.

- Right-hand limit as $x$ approaches $1$ from the right:
$\lim_{x \to 1^+} g(x) = \frac{3}{2} + 1 = \frac{5}{2}.$

Since the left-hand limit equals the right-hand limit and equals $g(1)$, $g(x)$ is continuous at $x = 1$.

Differentiability Check:

The derivative from the left side, $\frac{d}{dx} (\min\\{f(t)\\})$ at $x = 1$, does not match the derivative of $\frac{3}{2} + x$ from the right side.

Therefore, $g(x)$ is not differentiable at $x = 1$.

Thus, $g$ is continuous but not differentiable at $x = 1$.

The Correct answer is: g is continuous but not differentiable at x = 1