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Chapter 2: Problem 51

Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=|x+3| $$

Short Answer

Expert verified
The function \( y = |x + 3| \) is differentiable for all values of \( x \) except \( x = -3 \).

Step by step solution

01

Identify the kink point

Set the expression \( x + 3 \) inside the absolute value equal to zero. Solve for \( x \) to get the critical point where the function is not differentiable. So, \( x + 3 = 0 \) leads to \( x = -3 \).
02

Range of Differentiability

For a function to be differentiable, it must be continuous and have a single valued and non-vertical tangent line at every point in its domain. The absolute value function fails to meet this criterion at \( x = -3 \) because it has a kink at \( x = -3 \). So, the function is differentiable for all \( x\) values except \( x = -3 \).

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Key Concepts

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

Kink Point
A kink point is a specific location on a graph where the graph changes direction suddenly, creating a sharp corner. At these points, the slope of the tangent cannot be defined because the left-hand derivative and the right-hand derivative are not the same. For the function given in our exercise, \( y = |x + 3| \), a kink point occurs at \( x = -3 \).
  • The kink is created because the expression inside the absolute value \( x + 3 \) switches from positive to negative or vice versa at this point.
  • At this juncture, the slope is discontinuous, meaning the function cannot have a single defined tangent line.
  • This characteristic is crucial because it affects the function's differentiability.
Understanding the concept of a kink point requires acknowledging the idea that differentiability necessitates a smooth, unbroken curve in a graph, lacking any sharp angles or corners like the one at the kink point.
Absolute Value Function
The absolute value function transforms any input into its non-negative form. Mathematically, the expression is given by \( y = |x| \). In this scenario, for \( y = |x + 3| \), the graph shifts the basic absolute value graph three units to the left.
  • This shift occurs because every value of \( x \) in \( x + 3 \) is offset by adding 3.
  • The resulting graph is a "V" shaped curve opening upwards with its lowest point at \( x = -3 \).
  • Due to this V shape, there is a sharp turn at \( x = -3 \), making this function non-differentiable at that point.
The nature of absolute value functions, with their potential for causing sharp turns or corners, can often create challenges when determining differentiability, especially at those points where the internal expression equals zero.
Continuity
Continuity in functions is a foundational concept in calculus, essential for understanding differentiability. A function is continuous at a point when there is no break, jump, or hole in the graph at that location. For \( y = |x+3| \):
  • The function is continuous everywhere because absolute value functions do not have any breaks or jumps.
  • Even at \( x = -3 \), where we have a kink point, the function does not "break" but maintains continuity as the values approach and leave \(-3\).
  • This continuity is a necessary (but not solely sufficient) condition for differentiability.
While continuity ensures the graph is unbroken, differentiability requires that the slope be smoothly defined, which is not the case at our kink point \( x = -3 \).
Critical Point
Critical points in the realm of calculus are points on a function's graph where the derivative is zero or undefined. They are significant in analyzing the behavior of that function. With the function \( y = |x + 3| \), we identify \( x = -3 \) as a critical point.
  • At critical points like \( x = -3 \), the function is not differentiable, as the graph harbors a kink.
  • This is indicative because the derivative, which is the slope of the tangent line, cannot be consistently defined at these points.
  • While critical points often highlight maximum or minimum values or points of inflection, here it emphasizes a discontinuity in the derivative.
Understanding critical points allows us to discern not only maxima and minima but also to evaluate differentiability, highlighted by conditions like kink points, where the slope is undefined.

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Most popular questions from this chapter

Population Growth The population \(P\) (in thousands) of Japan can be modeled by \(P=-14.71 t^{2}+785.5 t+117,216\) where \(t\) is time in years, with \(t=0\) corresponding to 1980 . (a) Evaluate \(P\) for \(t=0,10,15,20,\) and \(25 .\) Explain these values. (b) Determine the population growth rate, \(d P / d t\) (c) Evaluate \(d P / d t\) for the same values as in part (a). Explain your results.

Marginal cost The cost \(C\) (in dollars) of producing \(x\) units of a product is given by \(C=3.6 \sqrt{x}+500 .\) (a) Find the additional cost when the production increases from 9 to 10 units. (b) Find the marginal cost when \(x=9 .\) (c) Compare the results of parts (a) and (b).

Use the table to answer the questions below. $$ \begin{array}{|cc|cc|}\hline \text { Quantity } & {} & {} & {} \\ {\text { produced }} & {} & {\text { Total }} & {\text { Marginal }} \\ {\text { and sold }} & {\text { Price }} & {(T R)} & {(M R)} \\ {(Q)} & {(p)} & {} & {(M R)} \\ \hline 0 & {160} & {0} & {-} \\ {2} & {140} & {280} & {130} \\ {4} & {120} & {480} & {90} \\ {6} & {100} & {600} & {50} \\ {8} & {80} & {640} & {10} \\ {10} & {60} & {600} & {-30} \\ \hline\end{array} $$ (a) Use the regression feature of a graphing utility to find a quadratic model that relates the total revenue \((T R)\) to the quantity produced and sold \((Q) .\) (b) Using derivatives, find a model for marginal revenue from the model you found in part (a). (c) Calculate the marginal revenue for all values of \(Q\) using your model in part (b), and compare these values with the actual values given. How good is your model?

Dow Jones Industrial Average The table shows the year-end closing prices \(p\) of the Dow Jones Industrial Average (DJIA) from 1992 through \(2006,\) where \(t\) is the year, and \(t=2\) corresponds to \(1992 .\) $$ \begin{array}{|c|c|c|c|c|c|}\hline t & {2} & {3} & {4} & {5} & {6} \\ \hline p & {3301.11} & {3754.09} & {3834.44} & {5117.12} & {6448.26} \\\ \hline\end{array} $$ $$ \begin{array}{|c|c|c|c|c|c|}\hline t & {7} & {8} & {9} & {10} & {11} \\\ \hline p & {7908.24} & {9181.43} & {11,497.12} & {10,786.85} & {10,021.50} \\\ \hline\end{array} $$ $$ \begin{array}{|c|c|c|c|c|c|}\hline t & {12} & {13} & {14} & {15} & {16} \\\ \hline p & {8341.63} & {10,453.92} & {10,783.01} & {10,717.50} & {12,463.15} \\\ \hline\end{array} $$ (a) Determine the average rate of change in the value of the DJIA from 1992 to 2006 . (b) Estimate the instantaneous rate of change in 1998 by finding the average rate of change from 1996 to 2000 . (c) Estimate the instantaneous rate of change in 1998 by finding the average rate of change from 1997 to 1999 . (d) Compare your answers for parts (b) and (c). Which interval do you think produced the best estimate for the instantaneous rate of change in \(1998 ?\)

Find \(f^{\prime}(x)\) $$ f(x)=\left(3 x^{2}-5 x\right)\left(x^{2}+2\right) $$
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