Problem 51 - Calculus - AP Edition

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}

Step-by-Step Solution

Answer

There exists an open interval around \(x_0\) where \(f\) is increasing since \(f^{\prime}(x_0) > 0\).

Step 1: Understand the Problem

You need to prove that in some open interval around \(x_0\), the function \(f\) is increasing, given that \(f\) is differentiable at \(x_0\) and \(f^{\prime}(x_0) > 0\).

Step 2: Use the Definition of Differentiability

Since \(f\) is differentiable at \(x_0\), the limit \( \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = f^{\prime}(x_0) \) exists and is positive.

Step 3: Apply the Limit Definition of Derivative

Since \(f^{\prime}(x_0) > 0\), there exists a \(\delta > 0\) such that for all \(x\) satisfying \(0 < |x - x_0| < \delta\), we have \(\frac{f(x) - f(x_0)}{x - x_0} > 0\).

Step 4: Choose an Open Interval

Let \(I = (x_0 - \delta, x_0 + \delta)\). This interval is open and contains \(x_0\). In this interval, for any \(x_1 < x_0 < x_2\), we have \(f^{\prime}(x_0) > 0\) implies \(f(x_1) < f(x_0) < f(x_2)\).

Step 5: Conclude the Result

From Steps 3 and 4, it follows that \(f\) is strictly increasing at points in the interval \(I\), establishing that \(f(x_1) < f(x_0) < f(x_2)\) for \(x_1 < x_0 < x_2\) chosen in \(I\).

Key Concepts

Derivative

The derivative of a function at a point provides crucial information about the behavior of the function around that point. If a function \( f \) is differentiable at \( x_0 \), it means that a tangent line with a definite slope can be drawn at that point.
This slope is precisely what the derivative measures. To be specific:

• The derivative \( f'\left(x_0\right) \) is a number representing the instantaneous rate of change of the function at \( x_0 \).
• This rate of change tells us whether the function is increasing or decreasing at \( x_0 \).

In our scenario, because \( f'\left(x_0\right) > 0 \), the function is increasing at this point. That means, around \( x_0 \), if we move slightly to the right or left, the function values will increase as we move to the right.
The positivity of the derivative makes this possible, indicating a rising trend in the function's values.

Open Interval

An open interval around a point on the number line does not include its endpoint values. Instead, it includes all the points between those endpoints. Open intervals are essential in calculus because they allow us to test properties of functions in a neighborhood around a specific point. In our exercise, the interval is pivotal:

• An open interval \( (x_0 - \delta, x_0 + \delta) \) indicates that no \( x \) within this range precisely equals \( x_0 \).
• This concept is crucial when analyzing limits or continuity since we only examine points arbitrarily close to \( x_0 \), but not necessarily at \( x_0 \).

This method is integral in showing that a function is increasing or decreasing in a region surrounding \( x_0 \). Such intervals enable us to make conclusions about the function's behavior without being directly at the tested point.

Increasing Function

An increasing function is one where larger inputs guarantee larger outputs. This means for a function \( f \) and any two points \( x_1 \) and \( x_2 \) such that \( x_1 < x_2 \), it must be that \( f(x_1) < f(x_2) \).

• This specific characteristic of the function holds true within any open interval where the derivative stays positive.
• If \( f'(x_0) > 0 \), as in our exercise, it signifies that in a sufficiently small interval around \( x_0 \), the function will consistently trend upwards.

So, if you choose any two points, \( x_1 < x_0 < x_2 \), both very close to \( x_0 \), then it’s guaranteed that \( f(x_1) < f(x_0) < f(x_2) \). This is how the concept of derivative directly relates to determining whether a function is increasing in a neighborhood around a point.