Problem 28
Question
Let \(f\) be continuous on \([a, b]\) and differentiable on \((a, b)\), and assume that \(f\) is not linear. Show that there are numbers \(s\) and \(t\) in \((a, b)\) such that \(f^{\prime}(s)>\frac{f(b)-f(a)}{b-a}\) and \(f^{\prime}(t)<\frac{f(b)-f(a)}{b-a}\) (Hint: Consider a point \((r, f(r))\) on the graph of \(f\) but not on the line joining \((a, f(a))\) and \((b, f(b))\). Use the Mean Value Theorem on \([a, r]\) and then on \([r, b] .\) )
Step-by-Step Solution
Verified Answer
There exist points \(s, t\) in \((a, b)\) such that \(f'(s)>m\) and \(f'(t)
1Step 1: Understanding the Problem
We are tasked with proving the existence of points \(s\) and \(t\) within the interval \((a, b)\) where the derivative \(f'(x)\) takes on values greater and lesser than the average rate of change on \([a, b]\), respectively.
2Step 2: Applying the Mean Value Theorem to \([a, r]\) and \([r, b]\)
According to the Mean Value Theorem (MVT), there exists a point \(c_1\) in \((a, r)\) such that \(f'(c_1) = \frac{f(r) - f(a)}{r - a}\) and a point \(c_2\) in \((r, b)\) such that \(f'(c_2) = \frac{f(b) - f(r)}{b - r}\).
3Step 3: Comparing Slopes and Derivatives
We denote the average rate of change on \([a, b]\) as \(m = \frac{f(b) - f(a)}{b - a}\). Since the function is not linear, it diverges from the line connecting \((a, f(a))\) and \((b, f(b))\) at some point \(r\). There will be at least one point in \((a, r)\) where the slope is greater than \(m\), and one in \((r, b)\) where it is less.
4Step 4: Conclusion from Existing Points
Since the function is continuous and differentiable, and it is not linear, the derivatives \(f'(c_1)\) and \(f'(c_2)\) ensure that there are points \(s\) and \(t\) in \((a, b)\) where \(f'(s) > m\) and \(f'(t) < m\). This follows the behavior expected from the Mean Value Theorem and the geometrical distribution of the slopes.
Key Concepts
ContinuityDifferentiabilityAverage Rate of ChangeNon-linear Functions
Continuity
Continuity is a fundamental property of functions that means the function has no abrupt jumps, holes, or breaks in its graph over the specified interval. For a function to be continuous on an interval \([a, b]\), it must be defined at every point within the interval, and for every point \(x\) in \([a, b]\), the limit of the function as \(x\) approaches that point must equal the function's value at that point.
- This ensures a smooth, unbroken path that the graph follows without lifting the pencil off the paper when drawing it.
- Continuity is crucial for applying the Mean Value Theorem (MVT) because it guarantees that there are no interruptions where the theorem cannot be applied.
In the context of the problem, the function \(f(x)\) being continuous on \([a, b]\) ensures that the function's values transition smoothly from \(f(a)\) to \(f(b)\), allowing us to analyze the behavior of the function across the entire interval.
- This ensures a smooth, unbroken path that the graph follows without lifting the pencil off the paper when drawing it.
- Continuity is crucial for applying the Mean Value Theorem (MVT) because it guarantees that there are no interruptions where the theorem cannot be applied.
In the context of the problem, the function \(f(x)\) being continuous on \([a, b]\) ensures that the function's values transition smoothly from \(f(a)\) to \(f(b)\), allowing us to analyze the behavior of the function across the entire interval.
Differentiability
Differentiability is the property of a function that allows it to have a derivative at a particular point or across an interval. When a function is differentiable on an interval \((a, b)\), it implies that the function's derivative exists at every point within that interval.
- Differentiability is a step further from continuity. A function can be continuous but not differentiable at a point (like a corner).
- However, if a function is differentiable, it is also continuous.
In our problem, because \(f(x)\) is differentiable on \((a, b)\), we are assured that we can calculate the derivative \(f'(x)\) at any point within that open interval. This property is crucial in proving the existence of the points s and t as required by the Mean Value Theorem, as it involves comparing these derivatives.
- Differentiability is a step further from continuity. A function can be continuous but not differentiable at a point (like a corner).
- However, if a function is differentiable, it is also continuous.
In our problem, because \(f(x)\) is differentiable on \((a, b)\), we are assured that we can calculate the derivative \(f'(x)\) at any point within that open interval. This property is crucial in proving the existence of the points s and t as required by the Mean Value Theorem, as it involves comparing these derivatives.
Average Rate of Change
The average rate of change of a function over an interval gives a measure of how much the function's output value changes about its input values. This is akin to finding the slope of the straight line connecting the starting point \(a\) to the endpoint \(b\) on the graph of the function. Mathematically, for a function \(f(x)\) over an interval \([a, b]\), it is calculated as:
\[\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}\]
- This value represents a linear approximation of the function's behavior between \(a\) and \(b\).
- It does not account for any curvature or nonlinear behavior between these points.
In our specific problem, the average rate of change provides a baseline for comparisons in the Mean Value Theorem. We are tasked to find derivative values \(f'(s)\) and \(f'(t)\) that are greater and less than this average rate, respectively, indicating the varying speed or direction of change of the function beyond a simple linear path.
\[\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}\]
- This value represents a linear approximation of the function's behavior between \(a\) and \(b\).
- It does not account for any curvature or nonlinear behavior between these points.
In our specific problem, the average rate of change provides a baseline for comparisons in the Mean Value Theorem. We are tasked to find derivative values \(f'(s)\) and \(f'(t)\) that are greater and less than this average rate, respectively, indicating the varying speed or direction of change of the function beyond a simple linear path.
Non-linear Functions
Non-linear functions are those whose graph is not a straight line, meaning they can curve, bend, twist, or loop.
- These functions can change their rate of increase or decrease and may have varying slopes at different points.
- Examples of non-linear functions include quadratics, exponentials, and trigonometrics.
In the context of the Mean Value Theorem, non-linear functions in \([a, b]\) play a pivotal role.
Given that \ f(x)\ is not linear in this problem setup, it tells us that the function does not simply connect the points \(a, f(a)\) and \(b, f(b)\) with a straight line.
This inherent non-linearity assures us that there exists at least a point \(r\) at which \(f(x)\), deviating from a straight path, exhibits different behavior whose derivative values fall above or below that of a constant line (average rate of change). This variation allows us to find the required points \(s\) and \(t\) where \(f'(s)\) is greater and \(f'(t)\) is lesser than the average rate of change.
- These functions can change their rate of increase or decrease and may have varying slopes at different points.
- Examples of non-linear functions include quadratics, exponentials, and trigonometrics.
In the context of the Mean Value Theorem, non-linear functions in \([a, b]\) play a pivotal role.
Given that \ f(x)\ is not linear in this problem setup, it tells us that the function does not simply connect the points \(a, f(a)\) and \(b, f(b)\) with a straight line.
This inherent non-linearity assures us that there exists at least a point \(r\) at which \(f(x)\), deviating from a straight path, exhibits different behavior whose derivative values fall above or below that of a constant line (average rate of change). This variation allows us to find the required points \(s\) and \(t\) where \(f'(s)\) is greater and \(f'(t)\) is lesser than the average rate of change.
Other exercises in this chapter
Problem 28
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