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Chapter 4: Problem 29

Determine whether the statement is true or false. Explain your answer. If the graph of \(f^{\prime}\) has a vertical asymptote at \(x=1\), then \(f\) cannot be continuous at \(x=1\).

Short Answer

Expert verified
False; a function can be continuous even if its derivative is not defined at that point.

Step by step solution

01

Understand the Problem Statement

The problem states that the derivative of a function, denoted as \(f'\), has a vertical asymptote at \(x=1\). We need to determine whether this implies that the original function \(f\) is not continuous at \(x=1\).
02

Understand Differentiability and Asymptotes

A vertical asymptote in \(f'\) at \(x=1\) indicates that \(f'\) becomes infinite or undefined at \(x=1\). If \(f'\) does not exist at \(x=1\), \(f\) is not differentiable at \(x=1\). However, a lack of differentiability doesn't automatically mean \(f\) is discontinuous at that point.
03

Explore Continuity and Differentiability

A function can be continuous at a point but not differentiable. For example, \(f(x) = |x|\) is continuous everywhere, including at \(x=0\), but is not differentiable at \(x=0\). This shows that continuity does not imply differentiability.
04

Conclusion with Counterexample

Consider the function \(f(x) = \sqrt{x}\) for \(x \geq 0\). Its derivative \(f'(x) = \frac{1}{2\sqrt{x}}\), which has a vertical asymptote at \(x=0\). Still, \(f(x) = \sqrt{x}\) is continuous at \(x=0\). Hence, the statement that \(f\) cannot be continuous at \(x=1\) is false.

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Key Concepts

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

Vertical Asymptote
A vertical asymptote occurs when a function approaches infinity as the input approaches some value. For the derivative of a function, denoted as \(f'(x)\), a vertical asymptote at \(x=1\) means that the derivative becomes infinitely large or undefined when \(x\) approaches 1. This characteristic signifies that the rate of change of the original function \(f(x)\) behaves very erratically near that point.
  • At a vertical asymptote, the graph of the function shoots upwards or downwards indefinitely.
  • It indicates a point of non-differentiability for the function \(f(x)\).
While a vertical asymptote in the derivative suggests a problem in differentiability, it does not always imply a lack of continuity for the original function.
Differentiability
Differentiability refers to the existence of a derivative at a particular point. For a function \(f(x)\) to be differentiable at \(x = c\), its derivative \(f'(x)\) must exist at that point.
  • A function is differentiable at a point if it has a well-defined tangent at that point.
  • However, if \(f'(x)\) has a vertical asymptote, the function \(f(x)\) is certainly not differentiable at this point.
It's important to remember that differentiability is a stronger condition than continuity. This means that even if a function is not differentiable at some point (like where \(f'(x)\) has a vertical asymptote), \(f(x)\) could still be continuous there. A classic example of such a function is the absolute value function, \(f(x) = |x|\), which is continuous everywhere but not differentiable at \(x=0\).
Counterexample
A counterexample serves to demonstrate that a given statement is not always true. In this scenario, the claim is that if \(f'(x)\) has a vertical asymptote at \(x=1\), then \(f(x)\) cannot be continuous at that point.
An effective counterexample provided is the function \(f(x) = \sqrt{x}\) for \(x \geq 0\).
  • Its derivative \(f'(x) = \frac{1}{2\sqrt{x}}\) has a vertical asymptote at \(x=0\), showing that the derivative becomes undefined there.
  • However, the function \(f(x) = \sqrt{x}\) is continuous at \(x=0\) since the limit of \(f(x)\) as \(x\) approaches 0 from the right is 0, which matches \(f(0)\).
This example shows that even with a vertical asymptote in the derivative, the original function can still remain continuous.
Graphical Analysis
Graphical analysis involves visualizing functions and their derivatives to understand concepts like vertical asymptotes and continuity. It helps to "see" what happens at particular critical points, providing greater insight beyond algebraic manipulation.
  • In the graph of \(f'(x)\), a vertical asymptote at \(x=1\) would appear as the graph shooting upwards or downwards indefinitely as it approaches the line \(x=1\).
  • The graph of \(f(x)\) could appear smooth and connected around \(x=1\), illustrating that \(f(x)\) remains continuous despite the asymptotic behavior of \(f'(x)\).
This graphical approach highlights the fact that continuity and differentiability are separate concepts. By analyzing graphs, students can more easily recognize that a function can have an "unruly" derivative without suffering from any discontinuity at the corresponding point.

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Most popular questions from this chapter

The logistic growth model given in Formula (1) is equivalent to $$ y e^{k t}+A y=L e^{k t} $$ where \(y\) is the population at time \(t(t \geq 0)\) and \(A, k,\) and \(L\) are positive constants. Use implicit differentiation to verify that $$ \begin{array}{l} \frac{d y}{d t}=\frac{k}{L} y(L-y) \\ \frac{d^{2} y}{d t^{2}}=\frac{k^{2}}{L^{2}} y(L-y)(L-2 y) \end{array} $$

Use a CAS to graph \(f^{\prime}\) and \(f^{\prime \prime},\) and then use those graphs to estimate the \(x\) -coordinates of the relative extrema of f. Check that your estimates are consistent with the graph of \(f\). $$f(x)=\frac{\tan ^{-1}\left(x^{2}-x\right)}{x^{2}+4}$$

(a) Use the Mean-Value Theorem to show that if \(f\) is differentiable on an interval, and if \(\left|f^{\prime}(x)\right| \leq M\) for all values of \(x\) in the interval, then $$ |f(x)-f(y)| \leq M|x-y| $$ for all values of \(x\) and \(y\) in the interval. (b) Use the result in part (a) to show that $$ |\sin x-\sin y| \leq|x-y| $$ for all real values of \(x\) and \(y\)

Verify that the hypotheses of Rolle's Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$f(x)=\frac{1}{2} x-\sqrt{x} ;[0,4]$$

In each part, find functions \(f\) and \(g\) that are increasing on \((-\infty,+\infty)\) and for which \(f-g\) has the stated property. (a) \(f-g\) is decreasing on \((-\infty,+\infty)\) (b) \(f-g\) is constant on \((-\infty,+\infty)\) (c) \(f-g\) is increasing on \((-\infty,+\infty)\)
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