Question

Let $f(x)$ be a real-valued function such that $f'(x_0)=0$ for some $x_0 \in (0,1)$, and $f''(x) > 0$ for all $x \in (0,1)$. Then $f(x)$ has

• A. no local minimum in (0,1)
• B. one local maximum in (0,1)
• C. exactly one local minimum in (0,1)
• D. two distinct local minima in (0,1)

Hint:

If $f''(x) > 0$ everywhere, the function is strictly convex and can have only one local minimum.

Answer 1

The Correct Option is C

Step 1: Use second derivative test.
Given $f'(x_0)=0$, so $x_0$ is a critical point. Since $f''(x) > 0$ for all $x \in (0,1)$, this implies the function is strictly convex throughout the interval.

Step 2: Effect of convexity.
A strictly convex function can have at most one critical point, and that point must be a local minimum. Therefore $x_0$ is the only local minimum.

Step 3: Conclusion.
$f(x)$ has exactly one local minimum inside $(0,1)$.