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Chapter 3: Problem 41

Determine if the piecewise-defined function is differentiable at the origin. $$f(x)=\left\\{\begin{array}{ll} 2 x-1, & x \geq 0 \\ x^{2}+2 x+7, & x<0 \end{array}\right.$$

Short Answer

Expert verified
The function is not differentiable at the origin because it is not continuous there.

Step by step solution

01

Check if the function is continuous at the origin

For a function to be differentiable at a point, it needs to be continuous there. To check continuity at the origin, ensure both pieces of the function give the same value at the point and as they approach it.- For the interval where \(x \geq 0\), evaluate \(f(x) = 2x - 1\) at \(x = 0\): \[f(0) = 2(0) - 1 = -1\]- For the interval where \(x < 0\), consider the limit as \(x\) approaches 0: \[\lim_{{x \to 0^-}} x^2 + 2x + 7 = 0^2 + 2(0) + 7 = 7\]These limits are not equal, hence the function is not continuous at \(x = 0\), and as a result, \(f(x)\) is not differentiable at \(x = 0\).
02

Conclude differentiability

Since continuity is violated at the origin, \(f(x)\) fails to be differentiable there. Differentiability requires both continuity and the existence of the derivative, meaning if a function is not continuous at a point, it cannot be differentiable there.

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Key Concepts

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

Piecewise Function
A piecewise function is a function composed of multiple sub-functions, each of which applies to a certain interval of the function's domain. The definition of the function changes depending on where the input falls within the specified intervals. To better understand:
  • Each "piece" of the function is defined with its own rule.
  • Typically requires separate calculations within each defined interval.
  • Boundaries are crucial to determine which piece applies.
Piecewise functions are common in mathematics because they allow for modeling real-world situations where a rule changes at certain thresholds. They are useful in describing situations with abrupt changes, like tax brackets or shipping costs.
Continuity
Continuity of a function at a specific point means that there is no interruption or jump in the output values as the input approaches that point. For a function to be continuous at a point, three conditions must be met:
  • The function is defined at the point, meaning there is a real number output.
  • The limit of the function as the input approaches the point exists.
  • The value of the function at the point is equal to this limit.
If any of these conditions fail, the function is not continuous at that point. For instance, in the provided exercise, at the origin, two different pieces of the function approach different values. This discrepancy results in discontinuity, which is a crucial factor in determining if the function is differentiable.
Limit
The concept of a limit is foundational in calculus and is primarily used to determine the value a function approaches as the input reaches a specific point. Limits help elucidate behavior at points where a function might not be explicitly defined. They are particularly useful in:
  • Assessing the behavior near discontinuities.
  • Evaluating derivatives, especially for functions defined with multiple rules.
  • Understanding asymptotic behavior or in infinite series.
For the function in the exercise, evaluating the limit as \(x\) approaches 0 from both sides (\(x \to 0^+\) and \(x \to 0^-\)) shows divergent values, indicating a break in continuity. This ultimately affects differentiability at such a point.

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Most popular questions from this chapter

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\sqrt{x}-\sin x, \quad[0,2 \pi], \quad a=2$$

Find \(y^{\prime \prime}\) $$y=\frac{1}{9} \cot (3 x-1)$$

Find \(d y / d t\) $$y=(t \tan t)^{10}$$

When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g.\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\). a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) to a new location. This increases the period by \(d T=0.001\) s. Find \(d g\) and estimate the value of \(g\) at the new location.

In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in W眉rzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about \(7 \mathrm{L} / \mathrm{min} .\) At rest it is likely to be a bit under \(6 \mathrm{L} / \mathrm{min}\). If you are a trained marathon runner running a marathon, your cardiac output can be as high as \(30 \mathrm{L} / \mathrm{min}\). Your cardiac output can be calculated with the formula $$y=\frac{Q}{D},$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\), \(y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}\), fairly close to the \(6 \mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?
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