Let f(x) be a real -valued function such that f'(x₀) = 0 for some x₀ ∈ (0, 1), and f"(x₀) > 0 for all x ∈ (0, 1). Then f(x) has

The correct option is: Exactly one local minimum in (0, 1).

To understand why this is correct, let us analyze the properties of the function based on the given conditions step-by-step.

Step 1: Identifying the critical point

We are given that there exists some point $x_0\in (0,1)$ such that $f\prime \left(x_0\right)=0$.

This tells us that $x_0$ is a critical point of the function $f\left(x\right)$ inside the interval $(0,1)$.

Step 2: Behavior of the second derivative

We are also given that the second derivative satisfies $f″\left(x\right)>0$ for all $x\in (0,1)$.

A strictly positive second derivative on an interval implies that:

1. The function $f\left(x\right)$ is strictly convex (concave upwards) on that interval.

2. The first derivative $f\prime \left(x\right)$ is a strictly increasing function on the interval $(0,1)$.

Step 3: Applying the Second Derivative Test

Since $f\prime \left(x_0\right)=0$ and $f″\left(x_0\right)>0$, by the second derivative test, $f\left(x\right)$ has a local minimum at $x=x_0$.

Step 4: Uniqueness of the local minimum

Because $f\prime \left(x\right)$ is strictly increasing on $(0,1)$, it can equal zero at most once in this interval.

Since $f\prime \left(x_0\right)=0$, there can be no other point in $(0,1)$ where the first derivative is zero.

Consequently, there are no other critical points where a local maximum or another local minimum could occur. Therefore, $f\left(x\right)$ has exactly one local minimum in the interval $(0,1)$.