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Chapter 2: Problem 51

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 \(x_{1}< x_{0}< x_{2},\) then \(f\left(x_{1}\right)< f\left(x_{0}\right)< f\left(x_{2}\right)\)

Short Answer

Expert verified
The function \(f\) is increasing around \(x_0\), so \(f(x_1) < f(x_0) < f(x_2)\) for points within an interval around \(x_0\).

Step by step solution

01

Understand the Problem

First, let's break down what it means for a function to be differentiable at a point and how the derivative relates to the function's behavior. If a function \(f\) is differentiable at \(x_0\), it means that the function is smooth and has a well-defined tangent at that point. The derivative \(f'(x_0) > 0\) tells us that the function is increasing at \(x_0\).
02

Apply the Definition of Derivative

The derivative at \(x_0\), \(f'(x_0)\), can be expressed as the limit \(\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}\). Since \(f'(x_0) > 0\), for sufficiently small positive \(h\), \(f(x_0 + h) > f(x_0)\), indicating that \(f\) is increasing to the right of \(x_0\). Conversely, if \(h\) is small and negative, \(f(x_0 + h) < f(x_0)\), showing the function is increasing to the left of \(x_0\) as well.
03

Construct the Open Interval

Given that \(f\) is increasing at \(x_0\), there exists an \(\epsilon > 0\) such that for any \(x\) in the interval \((x_0 - \epsilon, x_0 + \epsilon)\), \(f(x_0 - \delta) < f(x_0) < f(x_0 + \delta)\) for some \(\delta\) where \(-\epsilon < \delta < \epsilon\). This interval is where \(f(x)\) remains increasing.
04

Show the Behavior of \(f\) within the Interval

For any two points \(x_1\) and \(x_2\) within \((x_0 - \epsilon, x_0 + \epsilon)\) where \(x_1 < x_0 < x_2\), \(f(x_1) < f(x_0) < f(x_2)\). This inequality holds because \(f\) is continuously increasing within the interval, confirmed by the sign of the derivative.

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Key Concepts

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

Differentiable Function
A differentiable function is one that has a derivative at every point in its domain. This means the function is smooth, without any sharp corners or cusps, at each point. The key idea here is that differentiability guarantees not only the existence of a derivative, but also implies continuity. For a function to be differentiable, it must be able to form a tangent line at every point in its domain.
This concept is crucial when analyzing functions because it helps us understand how the function behaves and changes at different points. A differentiable function is predictable and well-behaved 鈥� it doesn鈥檛 do anything sudden or erratic around the points where it is differentiable. For example, for our function in the exercise, being differentiable at a point, like \( x_0 \), ensures that there's a smooth transition around this point and that the behavior (like increasing or decreasing) doesn't abruptly change.
Derivative
The derivative of a function at a particular point quantifies the rate at which the function's value changes. It's the slope of the tangent line to the function at that point. Mathematically, the derivative \( f'(x_0) \) can be thought of as the limit \( \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \). This tells us how the function's output changes with tiny, incremental changes in the input.
When we find a positive derivative, like \( f'(x_0) > 0 \) in the problem, it signifies that the function is increasing at \( x_0 \). This is because a positive slope leads upward as you move rightward along the x-axis. It's important to note that the derivative gives us localized information. If \( f'(x_0) > 0 \), we know that in the immediate vicinity around \( x_0 \), the function values increase.
Increasing Function
An increasing function is when the function's output gets larger as its input increases. In more formal terms, if \( x_1 < x_2 \) implies \( f(x_1) \le f(x_2) \) for all \( x_1 \) and \( x_2 \) in the domain of the function.
In the specific context of the problem, we consider a small interval around \( x_0 \), where \( f'(x_0) > 0 \). Given this condition, we find that the function increases continuously within the interval since the derivative is positive. This results in \( f(x_1) < f(x_0) < f(x_2) \) for points \( x_1 \) and \( x_2 \) around \( x_0 \) with \( x_1 < x_0 < x_2 \).
This increasing nature delineated by the derivative speaks to the rapid yet smooth growth of the function, allowing predictions about function behavior without graphing it.

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Most popular questions from this chapter

Find an equation for the tangent line to the graph at the specified value of \(x .\) $$ y=\tan \left(4 x^{2}\right), x=\sqrt{\pi} $$

Find an equation for the tangent line to the graph at the specified value of \(x .\) $$ y=\sec ^{3}\left(\frac{\pi}{2}-x\right), x=-\frac{\pi}{2} $$

If an object suspended from a spring is displaced vertically from its equilibrium position by a small amount and released, and if the air resistance and the mass of the spring are ignored, then the resulting oscillation of the object is called simple harmonic motion. Under appropriate conditions the displacement \(y\) from equilibrium in terms of time \(t\) is given by $$ y=A \cos \omega t $$ where \(A\) is the initial displacement at time \(t=0,\) and \(\omega\) is a constant that depends on the mass of the object and the stiffness of the spring (see the accompanying figure). The constant \(|A|\) is called the amplitude of the motion and \(\omega\) the angular frequency. (a) Show that $$ \frac{d^{2} y}{d t^{2}}=-\omega^{2} y $$ (b) The period \(T\) is the time required to make one complete oscillation. Show that \(T=2 \pi / \omega\). (c) The frequency \(f\) of the vibration is the number of oscillations per unit time. Find \(f\) in terms of the period \(T\) (d) Find the amplitude, period, and frequency of an object that is executing simple harmonic motion given by \(y=0.6 \cos 15 t,\) where \(t\) is in seconds and \(y\) is in centimeters.

(a) Use a graphing utility to obtain the graph of the function \(f(x)=x \sqrt{4-x^{2}}\) (b) Use the graph in part (a) to make a rough sketch of the graph of \(f^{\prime}\). (c) Find \(f^{\prime}(x),\) and then check your work in part (b) by using the graphing utility to obtain the graph of \(f^{\prime} .\) (d) Find the equation of the tangent line to the graph of \(f\) at \(x=1,\) and graph \(f\) and the tangent line together.

Find all values of \(a\) such that the curves \(y=a /(x-1)\) and \(y=x^{2}-2 x+1\) intersect at right angles.
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