Question

Which of the following functions \( f: \mathbb{R} \to \mathbb{R} \) has/have a local minimum at \( x = 0 \)?

• A. \( f(x) = \sin |x| \)
• B. \( f(x) = \sin x + \frac{x^3}{6} \)
• C. \( f(x) = x^4 + x^2 + 3 \)
• D. \( f(x) = \min \{x - \lfloor x \rfloor, 1 - x + \lfloor x \rfloor \} \)

Hint:

To determine if a function has a local minimum at a point, check for continuity, smoothness, and use the second derivative test when possible.

Answer 1

The Correct Option is A, C, D

Step 1: What a local minimum means.
At $x=0$ we want $f(0)$ to be the smallest value in a small interval around $0$. We check each function.

Step 2: Option A, $f(x)=\sin|x|$.
For small $x$, $\sin|x| \approx |x| \ge 0$ and it is zero only at $x=0$. So $0$ is the least nearby value. This is a local minimum, like a sharp V shape. Option A works.

Step 3: Option B, $f(x)=\sin x + \tfrac{x^3}{6}$.
Here $f'(x)=\cos x + \tfrac{x^2}{2}$ and $f'(0)=1$, which is not zero. So $0$ is not even a turning point. No minimum here.

Step 4: Option C, $f(x)=x^4+x^2+3$.
This is $3$ plus two terms that are positive for $x\neq 0$. So $f(0)=3$ is the smallest value nearby. Option C works.

Step 5: Option D, distance to nearest integer.
$f(x)=\min\{x-\lfloor x\rfloor,\,1-x+\lfloor x\rfloor\}$ is the gap to the closest whole number. Near $0$ it equals $|x|$ for small $x$, so $f(0)=0$ is again the least value. Option D works.

Step 6: Conclusion.
So A, C and D all have a local minimum at $0$.
\[ \boxed{A,\ C,\ D} \]