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

Question

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

Prepp Team · Accepted Answer

Local Extrema Analysis Using Derivatives

We are given a real-valued function f(x) with specific properties related to its derivatives on the interval (0, 1).

Function Derivative Conditions

There exists a point x0 within the interval (0, 1) such that f'(x0) = 0. This identifies x0 as a critical point, a potential location for a local maximum or minimum.
The second derivative, f"(x), is strictly positive (f"(x) > 0) for all values of x within the interval (0, 1). This indicates the function is concave upwards on this interval.

Second Derivative Test Application

The Second Derivative Test is a standard calculus tool used to classify critical points:

If f'(c) = 0 and f''(c) > 0, the function f(x) has a local minimum at the point x = c.
If f'(c) = 0 and f''(c) < 0, the function f(x) has a local maximum at the point x = c.

In this specific problem, we know f'(x0) = 0. Furthermore, because f"(x) > 0 for all x in (0, 1), this inequality must also hold true for x0, meaning f''(x0) > 0.

By applying the Second Derivative Test, the conditions f'(x0) = 0 and f''(x0) > 0 definitively indicate that f(x) possesses a local minimum at x = x0.

One Local Minimum Guarantee

The condition that f"(x) > 0 holds for the entire interval (0, 1) has an important implication: the function f(x) is strictly concave upwards everywhere in this interval. This concavity also means that its first derivative, f'(x), is a strictly increasing function on (0, 1).

A function that is strictly increasing can only cross the x-axis (equal zero) at a single point. Since we are given that f'(x0) = 0, it follows that x0 is the unique point within the interval (0, 1) where the derivative equals zero.

Consequently, there can be only one local extremum (which we've established is a minimum) within the interval (0, 1).

Derivative Properties Summary

The given condition f'(x0) = 0 establishes x0 as a critical point.
The condition f"(x) > 0 confirms that the function is concave up, classifying the critical point x0 as a local minimum.
The fact that f"(x) > 0 applies across the whole interval guarantees that x0 is the only critical point, hence the only local minimum.

Therefore, the function f(x) has exactly one local minimum in (0, 1).