/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 The right-sided and left-sided d... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The right-sided and left-sided derivatives of a function at a point a are given by $$f_{+}^{\prime}(a)=\lim _{h \rightarrow 0^{+}} \frac{f(a+h)-f(a)}{h} \text { and } f_{-}^{\prime}(a)=\lim _{h \rightarrow 0^{-}} \frac{f(a+h)-f(a)}{h},$$respectively, provided these limits exist. The derivative \(f^{\prime}(a)\) exists if and only if \(f_{+}^{\prime}(a)=f_{-}^{\prime}(a)\). a. Sketch the following functions. b. Compute \(f_{+}^{\prime}(a)\) and \(f_{-}^{\prime}(a)\) at the given point \(a\). c. Is \(f\) continuous at a? Is \(f\) differentiable at \(a ?\) $$f(x)=|x-2| ; a=2$$

Short Answer

Expert verified
Answer: The function is continuous at \(x = 2\), but not differentiable.

Step by step solution

01

Sketch the function

Represent the function: $$f(x) = |x - 2|$$ The graph of the function is a V-shape, with the point (2, 0) as its vertex. To the left of point (2, 0), the function is a decreasing linear function, represented by \(f(x) = 2 - x\). To the right of point (2, 0), the function is an increasing linear function, represented by \(f(x) = x - 2\).
02

Compute left-sided derivative

We first compute the left-sided derivative at point \(a = 2\): $$f_{-}^{\prime}(a) = \lim_{h \rightarrow 0^{-}}\frac{f(a+h) - f(a)}{h}$$ Since \(f(x) = 2 - x\) for \(x < 2\), we have: $$f_{-}^{\prime}(2) = \lim_{h \rightarrow 0^{-}}\frac{(2 - (2 + h)) - (2 - 2)}{h} = \lim_{h \rightarrow 0^{-}}\frac{-h}{h} = \lim_{h \rightarrow 0^{-}}(-1) = -1$$
03

Compute right-sided derivative

Next, we compute the right-sided derivative at point \(a = 2\): $$f_{+}^{\prime}(a) = \lim_{h \rightarrow 0^{+}}\frac{f(a+h) - f(a)}{h}$$ Since \(f(x) = x - 2\) for \(x > 2\), we have: $$f_{+}^{\prime}(2) = \lim_{h \rightarrow 0^{+}}\frac{((2 + h) - 2) - (2 - 2)}{h} = \lim_{h \rightarrow 0^{+}}\frac{h}{h} = \lim_{h \rightarrow 0^{+}}(1) = 1$$
04

Check continuity and differentiability

Now we can answer if the function is continuous and differentiable at point \(a = 2\): a. Continuity The function \(f(x) = |x - 2|\) is continuous at \(x = 2\) since both the left-sided and right-sided limits exist and are equal, and also equal to the value of the function at \(a = 2\): $$\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{+}} f(x) = f(2) = 0$$ b. Differentiability The function \(f(x) = |x - 2|\) is not differentiable at \(x = 2\) since the left-sided and right-sided derivatives are not equal: $$f_{-}^{\prime}(2) = -1 \neq f_{+}^{\prime}(2) = 1$$ In conclusion, the function \(f(x) = |x - 2|\) is continuous at point \(x = 2\) but not differentiable at that point.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Continuity of Functions
Continuity is a fundamental concept in calculus that describes the smooth behaviour of a function at a particular point or over an interval. A function is said to be continuous at a point \( a \) if the following conditions are satisfied:
  • The function \( f(x) \) is defined at \( x = a \).
  • The limit of \( f(x) \) as \( x \) approaches \( a \) exists.
  • The limit of \( f(x) \) as \( x \) approaches \( a \) is equal to the function value at \( a \), i.e., \( \lim_{x \to a} f(x) = f(a) \).
In the exercise, the function \( f(x) = |x-2| \) is continuous at \( x = 2 \) because both the left-hand limit and the right-hand limit as \( x \) approaches 2 are equal to \( f(2) = 0 \). This aligns with our understanding: for a function to be continuous, you can draw its graph without lifting your pencil. This continuity at \( x = 2 \) is verified by showing that \( \lim_{x \to 2^{-}} f(x) = \lim_{x \to 2^{+}} f(x) = 0 \).
Differentiability at a Point
Differentiability is about the ability to compute a tangent at any given point on a graph of a function. A function \( f(x) \) is said to be differentiable at a point \( a \) if the limit defining the derivative exists at that point. Specifically, differentiability at \( a \) requires that both the left-hand and right-hand derivatives exist and are equal:
  • The left-hand derivative at \( a \): \( f_{-}^{\prime}(a) = \lim_{h \to 0^{-}} \frac{f(a+h) - f(a)}{h} \)
  • The right-hand derivative at \( a \): \( f_{+}^{\prime}(a) = \lim_{h \to 0^{+}} \frac{f(a+h) - f(a)}{h} \)
For \( f(x) = |x - 2| \), these derivatives are \( -1 \) from the left and \( 1 \) from the right at \( x = 2 \). Since \( f_{-}^{\prime}(2) eq f_{+}^{\prime}(2) \), the function is not differentiable at this point. This discrepancy occurs because the graph has a sharp corner at \( x = 2 \).
Limit of a Function
Understanding limits is crucial for grasping the concepts of continuity and differentiability. The limit of a function \( f(x) \) as \( x \) approaches a point \( a \) is the value that \( f(x) \) gets closer to as \( x \) moves near \( a \). Limits help to define instantaneous rates of change and slope of tangent lines. They are defined as:
  • The limit of \( f(x) \) as \( x \to a^- \) (from the left): \( \lim_{x \to a^-} f(x) \).
  • The limit of \( f(x) \) as \( x \to a^+ \) (from the right): \( \lim_{x \to a^+} f(x) \).
In practice, for the function \( f(x) = |x - 2| \) around \( x = 2 \), both the left and the right limits approach \( 0 \). This confirms the function is continuous there, but their importance becomes even more prominent when analyzing the differentiability. Limits allow us to detect points of discontinuity or where a function is not smooth, such as cusps or sharp corners shown in the graph of \( |x - 2| \). Knowing how to compute these limits is key for evaluating the behavior of functions as inputs change.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose \(f\) is differentiable on an interval containing \(a\) and \(b\), and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x}\), show that \(c=\sqrt{a b}\), the geometric mean of \(a\) and \(b\), for \(a > 0\) and \(b > 0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a > 0\) and \(b > 0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

A boat is towed toward a dock by a cable attached to a winch that stands 10 feet above the water level (see figure). Let \(\theta\) be the angle of elevation of the winch and let \(\ell\) be the length of the cable as the boat is towed toward the dock. a. Show that the rate of change of \(\theta\) with respect to \(\ell\) is $$\frac{d \theta}{d \ell}=\frac{-10}{\ell \sqrt{\ell^{2}-100}}$$ b. Compute \(\frac{d \theta}{d \ell}\) when \(\ell=50,20,\) and \(11 \mathrm{ft}\) c. Find \(\lim _{t \rightarrow 10^{+}} \frac{d \theta}{d \ell}\) and explain what happens as the last foot of cable is reeled in (note that the boat is at the dock when \(\ell=10\) ). d. It is evident from the figure that \(\theta\) increases as the boat is towed to the dock. Why, then, is \(d \theta / d \ell\) negative?

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

The Binomial Theorem states that for any positive integer \(n\) $$\begin{aligned}(a+b)^{n}=& a^{n}+n a^{n-1} b+\frac{n(n-1)}{2 \cdot 1} a^{n-2} b^{2} \\ &+\frac{n(n-1)(n-2)}{3 \cdot 2 \cdot 1} a^{n-3} b^{3}+\cdots+n a b^{n-1}+b^{n} \end{aligned}$$ Use this formula and the definition \(f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\) to show that \(\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},\) for any positive integer \(n\).

Derivative of \(u(x)^{v(x)}\) Use logarithmic differentiation to prove that $$\frac{d}{d x}\left(u(x)^{v(x)}\right)=u(x)^{v(x)}\left(\frac{d v}{d x} \ln u(x)+\frac{v(x)}{u(x)} \frac{d u}{d x}\right)$$.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.