Problem 51 - Calculus Early Transcendentals

Question

Suppose that a function \(f\) is differentiable at \(x_{0}\) and that \(f^{\prime}\left(x_{0}\right)>0 .\) Prove that there exists an open interval containing \(x_{0}\) such that if \(x_{1}\) and \(x_{2}\) are any two points in this interval with \(x_{1}< x_{0}< x_{2},\) then \(f\left(x_{1}\right)< f\left(x_{0}\right)< f\left(x_{2}\right)\)

Step-by-Step Solution

Answer

The function \(f\) is increasing around \(x_0\), so \(f(x_1) < f(x_0) < f(x_2)\) for points within an interval around \(x_0\).

Step 1: Understand the Problem

First, let's break down what it means for a function to be differentiable at a point and how the derivative relates to the function's behavior. If a function \(f\) is differentiable at \(x_0\), it means that the function is smooth and has a well-defined tangent at that point. The derivative \(f'(x_0) > 0\) tells us that the function is increasing at \(x_0\).

Step 2: Apply the Definition of Derivative

The derivative at \(x_0\), \(f'(x_0)\), can be expressed as the limit \(\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}\). Since \(f'(x_0) > 0\), for sufficiently small positive \(h\), \(f(x_0 + h) > f(x_0)\), indicating that \(f\) is increasing to the right of \(x_0\). Conversely, if \(h\) is small and negative, \(f(x_0 + h) < f(x_0)\), showing the function is increasing to the left of \(x_0\) as well.

Step 3: Construct the Open Interval

Given that \(f\) is increasing at \(x_0\), there exists an \(\epsilon > 0\) such that for any \(x\) in the interval \((x_0 - \epsilon, x_0 + \epsilon)\), \(f(x_0 - \delta) < f(x_0) < f(x_0 + \delta)\) for some \(\delta\) where \(-\epsilon < \delta < \epsilon\). This interval is where \(f(x)\) remains increasing.

Step 4: Show the Behavior of \(f\) within the Interval

For any two points \(x_1\) and \(x_2\) within \((x_0 - \epsilon, x_0 + \epsilon)\) where \(x_1 < x_0 < x_2\), \(f(x_1) < f(x_0) < f(x_2)\). This inequality holds because \(f\) is continuously increasing within the interval, confirmed by the sign of the derivative.

Key Concepts

Differentiable Function

A differentiable function is one that has a derivative at every point in its domain. This means the function is smooth, without any sharp corners or cusps, at each point. The key idea here is that differentiability guarantees not only the existence of a derivative, but also implies continuity. For a function to be differentiable, it must be able to form a tangent line at every point in its domain.
This concept is crucial when analyzing functions because it helps us understand how the function behaves and changes at different points. A differentiable function is predictable and well-behaved – it doesn’t do anything sudden or erratic around the points where it is differentiable. For example, for our function in the exercise, being differentiable at a point, like \( x_0 \), ensures that there's a smooth transition around this point and that the behavior (like increasing or decreasing) doesn't abruptly change.

Derivative

The derivative of a function at a particular point quantifies the rate at which the function's value changes. It's the slope of the tangent line to the function at that point. Mathematically, the derivative \( f'(x_0) \) can be thought of as the limit \( \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \). This tells us how the function's output changes with tiny, incremental changes in the input.
When we find a positive derivative, like \( f'(x_0) > 0 \) in the problem, it signifies that the function is increasing at \( x_0 \). This is because a positive slope leads upward as you move rightward along the x-axis. It's important to note that the derivative gives us localized information. If \( f'(x_0) > 0 \), we know that in the immediate vicinity around \( x_0 \), the function values increase.

Increasing Function

An increasing function is when the function's output gets larger as its input increases. In more formal terms, if \( x_1 < x_2 \) implies \( f(x_1) \le f(x_2) \) for all \( x_1 \) and \( x_2 \) in the domain of the function.
In the specific context of the problem, we consider a small interval around \( x_0 \), where \( f'(x_0) > 0 \). Given this condition, we find that the function increases continuously within the interval since the derivative is positive. This results in \( f(x_1) < f(x_0) < f(x_2) \) for points \( x_1 \) and \( x_2 \) around \( x_0 \) with \( x_1 < x_0 < x_2 \).
This increasing nature delineated by the derivative speaks to the rapid yet smooth growth of the function, allowing predictions about function behavior without graphing it.