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

The Correct Option is A

To determine the continuity and differentiability of the function \( g(x) \) at \( x = 1 \), let's analyze its definition:

1. First, consider the function \( f(x) = \frac{x}{2} + \frac{2}{x} \). The domain of \( f(x) \) is \( (0, 2) \). 
2. The function \( g(x) \) is defined piecewise:
   • For \( 0 < x \leq 1 \), \( g(x) = \min\{f(t)\} \) where \( 0 < t \leq x \).
   • For \( 1 < x < 2 \), \( g(x) = \frac{3}{2} + x \).
3. First, we need to find the minimum value of \( f(t) \) in the interval \( (0, x] \) for \( x \leq 1 \). Since \( f(t) = \frac{t}{2} + \frac{2}{t} \), we differentiate to find the critical points:
   • \( f'(t) = \frac{1}{2} - \frac{2}{t^2} \)
   • Set \( f'(t) = 0 \) to find: \(t = \sqrt{4} = 2\). However, \( t = 2 \) is outside the interval \( (0, x] \) for \( x \leq 1 \).
   • Check endpoints within interval \( t \in (0, x]: f(x) \rightarrow f(x) = \frac{x}{2} + \frac{2}{x} \).
   • As \( x \rightarrow 1 \), consider \( \min\{ f(t) \} \) which simplifies to \( \frac{1}{2} + 2 = \frac{5}{2} \).
4. Therefore, \( g(x) \) for \( 0 < x \leq 1 \) is constant at \( \frac{5}{2} \).
5. For \( 1 < x < 2 \), \( g(x) = \frac{3}{2} + x \).
6. Check continuity at \( x = 1 \):
   • Left limit \( \lim_{x \to 1^-} g(x) = \frac{5}{2} \); as \( g(x) = \frac{5}{2} \) for \( x \leq 1 \).
   • Right limit \( \lim_{x \to 1^+} g(x) = \frac{3}{2} + 1 = \frac{5}{2} \); since \( g(x) = \frac{3}{2} + x \).
   • The left and right limits are equal, \( \frac{5}{2} \), which matches the value of \( g(x) \) at \( x = 1 \).
   • Thus, \( g(x) \) is continuous at \( x = 1 \).
7. Check differentiability at \( x = 1 \):
   • Since \( g(x) \) has different expressions on either side of \( x = 1 \), it is not differentiable at \( x = 1 \) as one of these does not have a derivative on this point.
   • Any slight change in \( x \) across \( x = 1 \) results in a different expression and slope of \( g(x) \).
8. Conclusion: \( g(x) \) is continuous but not differentiable at \( x = 1 \).

Thus, the correct answer is that \( g \) 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