/*! 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 49 Show that the function is contin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that the function is continuous but not differentiable at the given value of \(x\). \(f(x)=\left\\{\begin{array}{ll}x+2 & \text { if } x \leq 0 \\ 2-3 x & \text { if } x>0\end{array} ; x=0\right.\)

Short Answer

Expert verified
The function \(f(x)\) is continuous at \(x=0\) as both left and right limits are equal: \(\lim_{x \to 0^-} (x+2) = \lim_{x \to 0^+} (2-3x) = 2\). However, it is not differentiable at \(x=0\) since the left and right derivatives do not match: \(f'(0^-)=1 \neq f'(0^+)=-3\).

Step by step solution

01

Check limits from left and right at \(x=0\)

First, let's find the limits of the function at \(x=0\) to see if it is continuous: 1. For \(x \leq 0\), the function is defined as \(f(x)=x+2\). Let's find the left limit: \[ \lim_{x \to 0^-} (x+2) \] 2. For \(x>0\), the function is defined as \(f(x)=2-3x\). Let's find the right limit: \[ \lim_{x \to 0^+} (2-3x) \]
02

Evaluate left and right limits

Since the functions \(x+2\) and \(2-3x\) are continuous functions, we can just evaluate them at \(x=0\) to get their limits: 1. Left limit: \[ \lim_{x \to 0^-} (x+2) = (0+2) = 2 \] 2. Right limit: \[ \lim_{x \to 0^+} (2-3x) = (2-3(0)) = 2 \]
03

Check continuity

Since the left limit and the right limit are equal, the function is continuous at \(x=0\). In other words, \[ \lim_{x \to 0} f(x) = 2 \]
04

Examine differentiability

Now let's check the differentiability of the function at \(x=0\). 1. For \(x \leq 0\), the function is defined as \(f(x)=x+2\). The first derivative is \(f'(x)=1\). So, the left derivative is \[ f'(0^-)=1 \] 2. For \(x>0\), the function is defined as \(f(x)=2-3x\). The first derivative is \(f'(x)=-3\). So, the right derivative is \[ f'(0^+)=-3 \]
05

Draw conclusion

Since the left and right derivatives are not equal at \(x=0\), the function is not differentiable at that point. To summarize, - The function \(f(x)\) is continuous at \(x=0\) because the left limit and the right limit are equal: \(\lim_{x \to 0} f(x) = 2\). - The function \(f(x)\) is not differentiable at \(x=0\) because the left and right derivatives are not equal: \(f'(0^-)=1 \neq f'(0^+)=-3\).

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.

Limits of a Function
Understanding the concept of limits is crucial when exploring the continuity and behavior of functions at certain points. A limit essentially answers the question: as we approach a certain value of x, what value does the function approach? Let's consider what happens as we approach the number 0 from both the left (\(x \to 0^-\)) and the right (\(x \to 0^+\)) for our function.

When we compute the limit from the left side, we use the expression that defines the function for values less than or equal to zero; in the provided exercise, this is the linear function \(f(x)=x+2\). Conversely, to find the limit from the right side, we use the definition of the function for values greater than zero; for our function, this is the linear expression \(f(x)=2-3x\). If these two limits are equal, as they are in our example where both are 2, it suggests that the function approaches the same value from both sides at \(x=0\). This is a core indicator that our function is continuous at that particular point.
Piecewise Functions
Our focus function is an example of a piecewise function, which is constructed from multiple segments, each defined over a certain interval. The function changes its formula depending on the value of x. In this case, the function \(f(x)\) is defined as \(x+2\) when \(x \leq 0\) and as \(2-3x\) when \(x>0\).

One key aspect of understanding piecewise functions is examining how these 'pieces' interact at their boundaries. In order to be continuous, a piecewise function must not have any breaks or jumps at these boundary points. In our example, the boundary point at \(x=0\) is smooth because the value of both 'pieces' is 2; this seamless transition ensures continuity. To enhance learning, visual aids such as graphing the function can greatly assist in grasping the behavior at the boundary points, where students can visually verify the continuous nature of the function at \(x=0\).
Derivatives
When examining differentiability, we delve into the concept of derivatives—the rate at which a function is changing at any point. If a function is differentiable at a point, it means it has a well-defined tangent with a certain slope at that point. For our function \(f(x)\), we differentiate both pieces where \(f'(x)\) represents the derivative of \(f\).

The derivative \(f'(x)\) within the interval \(x \leq 0\) is a constant 1, indicating a steady, unchanging rate. However, the derivative for \(x > 0\) is -3, implying a different rate of change. At the boundary between these intervals, \(x=0\), if the rates do not match, we encounter what's called a 'corner' or 'cusp' in the graph of the function, indicating a point of non-differentiability. This is precisely what happens in our exercise, where the left and right derivatives do not match, resulting in a function that is continuous but not differentiable at \(x=0\). Using illustrative graphs to show the different slopes on each side can be especially helpful for students to visualize why there's no single tangent line at \(x=0\).

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

Find an equation of the tangent line to the given curve at the indicated point. $$ \frac{x^{2}}{9}-\frac{y^{2}}{4}=1 ; \quad\left(5, \frac{8}{3}\right) $$

The curve with equation \(x^{2 / 3}+y^{2 / 3}=4\) is called an astroid. Find an equation of the tangent line to the curve at the point \((3 \sqrt{3}, 1)\)

The Advance Visual Systems Corporation realizes a total profit of $$P(x)=-0.000002 x^{3}+0.016 x^{2}+80 x-70,000$$ dollars per week from the manufacture and sale of \(x\) units of their 26-in. LCD HDTVs. a. Find the marginal profit function \(P^{\prime}\). b. Compute \(P^{\prime}(2000)\) and interpret your result.

A horizontal uniform beam of length \(L\) is supported at both ends and bends under its own weight \(w\) per unit length. Because of its elasticity, the beam is distorted in shape, and the resulting distorted axis of symmetry (shown dashed in the figure) is called the elastic curve. It can be shown that an equation for the elastic curve is $$y=\frac{w}{24 E I}\left(x^{4}-2 L x^{3}+L^{3} x\right)$$ where the product \(E I\) is a constant called the flexural rigidity. (a) The distorted beam (b) The elastic curve in the \(x y\) -plane (The positive direction of the \(y\) -axis is directed downward.) a. Find the angle that the elastic curve makes with the positive \(x\) -axis at each end of the beam in terms of \(w, E\), and \(I .\) b. Show that the angle that the elastic curve makes with the horizontal at \(x=L / 2\) is zero. c. Find the deflection of the beam at \(x=L / 2\). (We will show that the deflection is maximal in Section 3.1, Exercise 74.)

Use the result of Exercise 112 to find the derivative of the function. $$ g(x)=x\left|x^{2}+x\right| $$

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.