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Chapter 2: Problem 87

Use a graphing utility to find the \(x\) -values at which \(f\) is differentiable. \(f(x)=|x+3|\)

Short Answer

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

Step by step solution

01

Understand the function

The function \(f(x) = |x+3|\) is an absolute value function. It's a piecewise-defined function and can be expressed as \(f(x) = -(x+3)\) if \(x < -3\), and \(f(x) = (x+3)\) if \(x > -3\).
02

Identify where the function is differentiable

A function is differentiable at 'x' if it has a well-defined and non-discontinuous derivative at that point. For the absolute function, there is a sharp turn, or cusp, at \(x = -3\). Therefore, this is the point where the function is not differentiable.
03

Use the graphing utility

Using a graphing utility, plot \(f(x) = |x+3|\). The graph will confirm that the function has a cusp at \(x = -3\). Hence, the function is differentiable for all values of 'x' except at \(x = -3\).

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Key Concepts

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

Understanding the Absolute Value Function
The absolute value function is a key concept in mathematics that often appears in various contexts. When you see a function like \(f(x) = |x+3|\), it represents the distance of the expression inside the absolute value from zero on the number line. This means it will always produce a non-negative result.

For any real number, \(x\), the absolute value function is defined as follows:
  • If \(x + 3\) is positive, the output is simply \(x + 3\).
  • If \(x + 3\) is negative, the output is the negation of \(x + 3\), which results in \(-(x+3)\).

In simpler terms, it flips negative inputs into positive outputs. This characteristic leads to the creation of a V-shaped graph if plotted.
Piecewise-Defined Function
The concept of a piecewise-defined function arises when a function is defined by different expressions over different parts of its domain. For the function \(f(x) = |x+3|\), it can be broken down into two linear components based on the value of \(x\):
  • For \(x < -3\), the expression becomes \(f(x) = -(x+3)\).
  • For \(x > -3\), it changes to \(f(x) = x+3\).

This piecewise nature is necessary because of the change in behavior at the point \(x = -3\), giving it two distinct linear parts.
Utilizing a Graphing Utility
A graphing utility is a great tool for visualizing mathematical functions. It allows you to input functions and see their graphical representation, which is particularly helpful when dealing with functions that are not simple straight lines.

For \(f(x) = |x+3|\), a graphing utility will show a V-shape graph. This graph visually confirms the piecewise nature of the function and where certain characteristics like sharp turns occur. By analyzing the graph, you can better understand where the function is differentiable, as well as other key points where the behavior of the function changes dramatically.
What is a Sharp Turn or Cusp?
When we talk about differentiability, we also need to consider points where a function might not be differentiable. One such scenario is a sharp turn or cusp. In the case of \(f(x) = |x+3|\), a cusp occurs at \(x = -3\), where the function abruptly changes direction without a smooth transition.

A cusp indicates a point of non-differentiability because the derivative, which represents the slope of the tangent, is not well-defined. At a point like \(x = -3\), you would notice that the slope changes abruptly, resulting in no single tangent line that can smoothly connect both sides.

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Most popular questions from this chapter

A buoy oscillates in simple harmonic motion \(y=A \cos \omega t\) as waves move past it. The buoy moves a total of \(3.5\) feet (vertically) from its low point to its high point. It returns to its high point every 10 seconds. (a) Write an equation describing the motion of the buoy if it is at its high point at \(t=0\). (b) Determine the velocity of the buoy as a function of \(t\).

(a) Use implicit differentiation to find an equation of the tangent line to the ellipse \(\frac{x^{2}}{2}+\frac{y^{2}}{8}=1\) at \((1,2)\). (b) Show that the equation of the tangent line to the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) at \(\left(x_{0}, y_{0}\right)\) is \(\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1\)

The cost of producing \(x\) units of a product is \(C=60 x+1350\). For one week management determined the number of units produced at the end of \(t\) hours during an eight-hour shift. The average values of \(x\) for the week are shown in the table. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \boldsymbol{t} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \boldsymbol{x} & 0 & 16 & 60 & 130 & 205 & 271 & 336 & 384 & 392 \\ \hline \end{array} $$ (a) Use a graphing utility to fit a cubic model to the data. (b) Use the Chain Rule to find \(d C / d t\). (c) Explain why the cost function is not increasing at a constant rate during the 8 -hour shift.

Verify that the two families of curves are orthogonal where \(C\) and \(K\) are real numbers. Use a graphing utility to graph the two families for two values of \(C\) and two values of \(K\). $$ x^{2}+y^{2}=C^{2}, \quad y=K x $$

Use a graphing utility to sketch the intersecting graphs of the equations and show that they are orthogonal. [Two graphs are orthogonal if at their point(s) of intersection their tangent lines are perpendicular to each other.] $$ \begin{aligned} &2 x^{2}+y^{2}=6 \\ &y^{2}=4 x \end{aligned} $$
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