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Chapter 5: Problem 46

Assume that \(f\) is continuous on \([0,1]\) and differentiable on \((0,1)\). Assume that \(f^{\prime}(1 / 2)=0\), show by sketching the graph of a function \(f(x)\) that satisfies all of these conditions (you do not need to write down the equation of the function) that it is not necessary that \(f(0)=f(1)\).

Short Answer

Expert verified
A function can meet all these conditions and still have \( f(0) \neq f(1) \); consider a concave up or down parabola touching horizontal at x=0.5.

Step by step solution

01

Understand the Problem

We need to visualize a function \( f(x) \) that is continuous from 0 to 1, differentiable between 0 and 1, and has a derivative of zero at \( x = \frac{1}{2} \). We also need to show that despite these conditions, \( f(0) \) and \( f(1) \) do not have to be equal.
02

Concept of Continuity

Continuity on \([0, 1]\) means that there are no breaks or jumps in the graph of \( f(x) \) from 0 to 1. Imagine a smooth line that can be drawn without lifting the pen from paper that connects the point \( f(0) \) to \( f(1) \).
03

Concept of Differentiability

The condition that \( f \) is differentiable on \((0, 1)\) indicates it has a well-defined tangent at every point between 0 and 1, making the graph smooth, with no sharp corners.
04

Interpretation of Derivative Condition

The condition \( f'(\frac{1}{2}) = 0 \) implies that \( f(x) \) has a horizontal tangent line at \( x = \frac{1}{2} \). This means there is a local maximum or minimum at this point so the curve flattens out at \( x = \frac{1}{2}\).
05

Sketching Example

To visualize this, imagine a curved line, like a hill, where the peak or trough is exactly at \( x = \frac{1}{2} \). Depending on the height of the curve at the endpoints, it could start at a higher point and end at a lower point (or vice versa), illustrating \( f(0) eq f(1) \). For instance, a downward-opening parabola like \( f(x) = -(x - \frac{1}{2})^2 + \frac{1}{4} \) satisfies the criteria but has \( f(0) < f(1) \).
06

Conclusion

By constructing a function that meets all the conditions but has different values for \( f(0) \) and \( f(1) \), we demonstrate that continuity, differentiability, and a point at which the derivative is zero do not necessarily imply equal function values at the endpoints.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity
Continuity is a concept in calculus that ensures there are no breaks, jumps, or holes in the graph of a function over a given interval. For a function to be continuous on a closed interval \([a, b]\), it must be defined at every point within the interval and must not break when plotting between any two points. This is akin to drawing a line on paper without ever lifting your pen. In the context of our exercise, the function \( f(x) \) must be continuous from 0 to 1. This ensures a smooth transition in the values of the function from \( f(0) \) to \( f(1) \). However, this does not mean the function needs to have the same value at both \( f(0) \) and \( f(1) \). Think of \( f(x) \) as a bridge, spanning the gap between these two points, yet the heights (function values) at each endpoint can differ.
Differentiability
In calculus, differentiability entails that a function has a defined derivative at every point within a given interval. This implies that the function has a well-defined tangent line at every point; in simpler terms, the graph of the function is smooth, without any sharp corners or cusps. For our particular problem, the function \( f(x) \) is differentiable over the open interval \( (0, 1) \). This means you can slide your finger gently over the curve between these points without encountering any abrupt changes in direction. Differentiability guarantees that the slope of \( f(x) \) can be calculated everywhere on this interval. However, differentiability does not automatically impose equal values at the endpoints of \( f \); it merely assures the consistency of the function's smoothness across the specified interval.
Derivative
The derivative of a function at a specific point is the slope of the tangent line to the function's graph at that point. It provides information on how the function is changing at that particular location. In our exercise, knowing that \( f'\left(\frac{1}{2}\right) = 0 \) tells us that there is a horizontal tangent line, indicating a local maximum or minimum at \( x = \frac{1}{2} \). This implies that the rate of change of the function at this point is zero; essentially, the function flattens out here. The derivative's significance in this exercise is that it describes the behavior of the function at this critical midpoint, but it still allows for the values at the limits \( f(0) \) and \( f(1) \) to differ, providing a peak or trough in the graph centered at \( x = \frac{1}{2} \).
Function Sketching
Function sketching is a powerful tool in understanding the qualitative behavior of a function. By sketching a graph of \( f(x) \), you can visualize how it behaves across an interval. Based on the conditions given in the problem, our sketch might resemble a gentle curve with a peak or trough at \( x = \frac{1}{2} \), such as the visualization of a hill or valley. Consider a parabolic graph that opens downwards representing \( f(x) = -\left(x - \frac{1}{2}\right)^2 + \frac{1}{4} \). This sketch comports with a local maximum at \( x = \frac{1}{2} \) and emphasizes that while the curve is continuous and differentiable from 0 to 1, the values at the endpoints can indeed be different. Visualization not only helps in solving the problem at hand but also deepens the understanding of how the concepts of continuity, differentiability, and derivatives interplay to form the shape and properties of a function's graph.

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Most popular questions from this chapter

In Problem 9 we neglected to consider the time delay between a pill being taken and the drug entering the patient's blood. In Chapter 8 we will introduce compartment models as models for drug absorption. We will show that a good model for a drug being absorbed from the gut is that the rate of drug absorption, \(A(t)\), varies with time according to: $$ A(t)=C e^{-k t}, t \geq 0 $$ where \(C>0\) and \(k>0\) are coefficients that will depend on the type of drug, as well as varying between patients. (a) Assume that the drug has first order elimination kinetics, with elimination rate \(k_{1} .\) Show that the amount of drug in the patient's blood will obey a differential equation: $$ \frac{d M}{d t}=C e^{-k t}-k_{1} M $$ (b) Verify that a solution of this differential equation is: $$ M(t)=\frac{C e^{-k t}}{k_{1}-k}+a e^{-k_{1} t} $$ where \(a\) is any coefficient, and we assume \(k_{1} \neq k\). (c) To determine the coefficient \(a\), we need to apply an initial condition. Assume that there was no drug present in the patient's blood when the pill first entered the gut (that is, \(M(0)=0\) ). Find the value of \(a\). (d) Let's assume some specific parameter values. Let \(C=2\), \(k=3\), and \(k_{1}=1 .\) Show that \(M(t)\) is initially increasing, and then starts to decrease. Find the maximum level of drug in the patient's blood. (e) Show that \(M(t) \rightarrow 0\) as \(t \rightarrow \infty\). (f) Using the information from (d) and (e), make a sketch of \(M(t)\) as a function of \(t\).

Find the general antiderivative of the given function. $$ f(x)=\sec ^{2}(2 x) $$

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