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)

Let's analyze the properties of the function step-by-step:

We are given that the second derivative of the real-valued function $f\left(x\right)$ satisfies:
$f^{″}\left(x\right)>0$
for all $x\in (0,1)$.

Since the second derivative is strictly positive on the interval $(0,1)$, the first derivative $f^{\prime }\left(x\right)$ must be a strictly increasing function on $(0,1)$.

We are also given that there exists a point $x_0\in (0,1)$ such that:
$f^{\prime }\left(x_0\right)=0$

Because $f^{\prime }\left(x\right)$ is strictly increasing on $(0,1)$:
1. For any $x<x_0$ in the interval, we must have $f^{\prime }\left(x\right)<f^{\prime }\left(x_0\right)=0$.
2. For any $x>x_0$ in the interval, we must have $f^{\prime }\left(x\right)>f^{\prime }\left(x_0\right)=0$.

This tells us that:
- The function $f\left(x\right)$ is strictly decreasing on the subinterval $(0,x_0]$ because its derivative is negative.
- The function $f\left(x\right)$ is strictly increasing on the subinterval $[x_0,1)$ because its derivative is positive.

By the first derivative test, since the derivative changes sign from negative to positive at $x_0$, the function $f\left(x\right)$ has a local minimum at $x=x_0$.

Furthermore, since $f^{\prime }\left(x\right)$ is strictly increasing, it can cross zero at most once in $(0,1)$. Thus, $x_0$ is the unique critical point of $f\left(x\right)$ in the interval. Consequently, there are no other local extrema (maximum or minimum) in $(0,1)$.

Therefore, the function $f\left(x\right)$ has exactly one local minimum in the interval $(0,1)$.