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Prove that if \(f\) has a derivative at \(p\), then \(f(p)\) is finite, provided \(f\) is not constantly infinite on any interval \((p, q)\) or \((q, p), p \neq q\). [Hint: If \(f(p)=\pm \infty\), each \(G_{p}\) has points at which \(\frac{\Delta f}{\Delta x}=+\infty\), as well as those \(x\) with \(\frac{\Delta f}{\Delta x}=-\infty .\) ]

Short Answer

Expert verified
If \( f \) has a derivative at \( p \), then \( f(p) \) is finite because otherwise, the derivative limit would not be finite.

Step by step solution

01

Understand the Assumption of Derivative

If \( f \) has a derivative at \( p \), it means that the limit \( \lim_{x \to p} \frac{f(x) - f(p)}{x - p} \) exists and is finite. This limit represents the slope of the tangent line to the graph of \( f \) at the point \( x = p \).
02

Assume the Derivative Definition

By definition, the derivative \( f'(p) \) is given by \( \lim_{x \to p} \frac{f(x) - f(p)}{x - p} = L \), where \( L \) is a finite number.
03

Address the Hint with the Given Conditions

Assume for the sake of contradiction that \( f(p) = \pm \infty \). If \( f(p) = +\infty \), then, for any \( \epsilon > 0 \), the slope \( \frac{f(x) - f(p)}{x - p} \) becomes \( +\infty \) or approaches it as \( x \to p \), leading to a contradiction with the existence of a finite limit \( f'(p) \). Similarly, if \( f(p) = -\infty \), \( \frac{f(x) - f(p)}{x - p} \) would become \( -\infty \).
04

Conclude Using the Contradiction

Since both scenarios lead to \( \frac{f(x) - f(p)}{x - p} \) becoming infinite as \( x \to p \), and since we have a finite limit \( L \) by assumption, \( f(p) \) cannot be infinite. Hence, \( f(p) \) must be finite.

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Key Concepts

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

Finite Limits
Finite limits are a foundational concept in calculus, particularly when discussing the behavior of functions as they approach certain points. In simple terms, a finite limit indicates that as the input of a function nears a specific value, the output of the function approaches a particular, fixed number. This fixed number represents the limit, and it is crucial for determining the continuity and differentiability of functions.
In the context of derivatives, the existence of a finite limit is essential. Consider the limit \( \lim_{x \to p} \frac{f(x) - f(p)}{x - p} \). For a derivative to exist at \( p \), this expression must result in a finite number. This confirms that the change in the function \( f(x) \) with respect to \( x \) is predictable and does not jump to infinity.
By ensuring finite limits, mathematicians can analyze functions and predict behavior in a controlled environment, which is integral to many applications in science and engineering. When limits are finite, we ensure the function behaves "well" at that particular point, without wild oscillations.
Tangent Line
The tangent line at a given point on a curve provides a linear approximation of the curve near that point. It touches the curve only at the specific point in question and represents the instant rate of change of the function at that position.
Visualize a curve on a graph. The slope of the tangent line at any point \( p \) on this curve is represented by the derivative \( f'(p) \). This slope is calculated by the limit \( \lim_{x \to p} \frac{f(x) - f(p)}{x - p} \).
The tangent line is significant because it gives us the best linear approximation of the function near \( p \). In everyday terms, it tells us how steeply the function is increasing or decreasing at that point. Metaphorically, it's like using a ruler to touch a curve at one point and observe how the curve behaves right around that contact point.
Infinite Limits
Infinite limits occur when the output of a function grows without bounds as the input approaches a certain value. In other words, as \( x \) gets very close to a particular point, \( f(x) \) increases or decreases indefinitely.
Mathematically, an infinite limit is expressed as \( \lim_{x \to p} f(x) = \pm \infty \). This happens when the function does not settle at a finite value but instead heads off towards infinity. Infinite limits can complicate analysis because they suggest discontinuity or non-differentiability at the particular point.
When examining derivatives, having an infinite limit as \( x \to p \) would imply the slope of the tangent line goes infinite. This contradicts the idea of a finite derivative, as derivatives must represent a real number reflecting the slope of a tangent line.
Contradiction in Calculus
A contradiction in calculus often arises when an assumption leads to an impossible conclusion, causing the initial assumption to be false. This technique, known as proof by contradiction, is a powerful tool in mathematics.
In the given exercise, we started by assuming \( f(p) = \pm \infty \), leading us to explore the limits of \( \frac{f(x) - f(p)}{x - p} \) as \( x \to p \). Under this assumption, the limit would be infinite, either \( +\infty \) or \( -\infty \). However, for a function to have a derivative at \( p \), this limit has to be finite.
This infinite outcome contradicts our initial condition that the derivative exists and is finite, thus proving the assumption \( f(p) = \pm \infty \) to be incorrect. Therefore, \( f(p) \) must be finite, as deduced from the contradiction. This method is elegant and clear, often used in mathematical proofs to show that certain conditions must or must not hold.

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Most popular questions from this chapter

In the following cases show that \(V_{f}[I]=+\infty,\) though \(f\) is bounded on I. (In case (iii), \(f\) is continuous, and in case (iv), it is even differentiable on \(I .)\) $$ \text { (i) For } I=[a, b](a

Prove that if the functions \(f_{i}: E^{1} \rightarrow E^{*}(C)\) are differentiable at \(p,\) so is their product, and $$ \left(f_{1} f_{2} \cdots f_{m}\right)^{\prime}=\sum_{i=1}^{m}\left(f_{1} f_{2} \cdots f_{i-1} f_{i}^{\prime} f_{i+1} \cdots f_{m}\right) \text { at } p $$

Let $$ f(x)=\sin x \text { and } g(x)=\cos x $$ Show that \(f\) and \(g\) are differentiable on \(E^{1},\) with $$ f^{\prime}(p)=\cos p \text { and } g^{\prime}(p)=-\sin p \text { for each } p \in E^{1} . $$ Hence prove for \(n=0,1,2, \ldots\) that $$ f^{(n)}(p)=\sin \left(p+\frac{n \pi}{2}\right) \text { and } g^{(n)}(p)=\cos \left(p+\frac{n \pi}{2}\right) . $$

Prove that if \(\int f\) exists on each \(I_{n}=\left[a_{n}, b_{n}\right],\) where $$ a_{n+1} \leq a_{n} \leq b_{n} \leq b_{n+1}, \quad n=1,2, \ldots $$ then \(\int f\) exists on $$ I=\bigcup_{n=1}^{\infty}\left[a_{n}, b_{n}\right] $$ itself an interval with endpoints \(a=\inf a_{n}\) and \(b=\sup b_{n}, a, b \in E^{*}\). [Hint: Fix some \(c \in I_{1}\). Define $$ H_{n}(t)=\int_{c}^{t} f \text { on } I_{n}, n=1,2, \ldots $$ Prove that $$ (\forall n \leq m) \quad H_{n}=H_{m} \text { on } I_{n}\left(\text { since }\left\\{I_{n}\right\\} \uparrow\right) . $$ Thus \(H_{n}(t)\) is the same for all \(n\) such that \(t \in I_{n},\) so we may simply write \(H\) for \(H_{n}\) on \(I=\bigcup_{n=1}^{\infty} I_{n} .\) Show that \(H=\int f\) on all of \(I ;\) verify that \(I\) is, indeed, an interval.]

Let \(I=[0,2 \pi]\) and define \(f, g, h: E^{1} \rightarrow E^{2}(C)\) by $$ \begin{aligned} f(x) &=(\sin x, \cos x) \\ g(x) &=(\sin 3 x, \cos 3 x) \\ h(x) &=\left(\sin \frac{1}{x}, \cos \frac{1}{x}\right) \text { with } h(0)=(0,1) . \end{aligned} $$ Show that \(f[I]=g[I]=h[I]\) (the unit circle; call it \(\mathrm{A}),\) yet \(\ell_{f} A=2 \pi\) \(\ell_{g} A=6 \pi,\) while \(V_{h}[I]=+\infty\). (Thus the result of Problem 7 fails for closed curves and nonsimple arcs.)

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