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Is there a value of \(b\) that will make $$g(x)=\left\\{\begin{array}{ll}x+b, & x<0 \\\\\cos x, & x \geq 0\end{array}\right.$$ continuous at \(x=0 ?\) Differentiable at \(x=0 ?\) Give reasons for your answers.

Short Answer

Expert verified
The function is continuous at \(x=0\) if \(b=1\), but it is not differentiable at \(x=0\).

Step by step solution

01

Understanding Continuity Conditions

For the function \(g(x)\) to be continuous at \(x = 0\), the left-hand limit \(\lim_{x \to 0^-} g(x)\), the right-hand limit \(\lim_{x \to 0^+} g(x)\), and \(g(0)\) must all be equal.
02

Calculate Left-Hand Limit

For \(x < 0\), \(g(x) = x + b\). As \(x\) approaches 0 from the left, \(\lim_{x \to 0^-} g(x) = b\).
03

Calculate Right-Hand Limit

For \(x \geq 0\), \(g(x) = \cos x\). As \(x\) approaches 0 from the right, \(\lim_{x \to 0^+} g(x) = \cos(0) = 1\).
04

Ensure Continuity at x=0

Set the left-hand and right-hand limits equal to each other to ensure continuity: \(b = 1\). Also, \(g(0) = \cos(0) = 1\), so the function is continuous at \(x = 0\) if \(b = 1\).
05

Understanding Differentiability Conditions

For \(g(x)\) to be differentiable at \(x = 0\), the derivative from the left must equal the derivative from the right at that point.
06

Calculate Left-Hand Derivative

For \(x < 0\), \(g(x) = x + b\). The derivative \(\frac{d}{dx}(x + b) = 1\).
07

Calculate Right-Hand Derivative

For \(x \geq 0\), \(g(x) = \cos x\). The derivative \(\frac{d}{dx}(\cos x) = -\sin x\). At \(x = 0\), \(-\sin(0) = 0\).
08

Compare Derivatives for Differentiability

At \(x = 0\), the left-hand derivative is 1 and the right-hand derivative is 0. Since they are not equal, \(g(x)\) is not differentiable at \(x = 0\).

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Key Concepts

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

Piecewise Functions
Piecewise functions are distinctive types of functions that have different expressions or rules for different parts of their domains. The graph of a piecewise function is often composed of separate pieces that connect under specific rules.
In mathematical notation, a piecewise function is typically written with a bracket and formulas for each segment of its domain. For example, the function in the exercise is given as:
  • When \(x < 0\), the function is expressed as \(x + b\).
  • When \(x \geq 0\), the function switches to a new rule, \(\cos x\).
This particular piecewise function changes its rule at \(x = 0\). The challenge with piecewise functions is ensuring continuity and differentiability at the points where the rules switch, especially at those transition points like \(x = 0\) in this question.
Limits
Limits are foundational concepts in calculus and are used to describe the behavior of functions as they approach a certain point. In examining a piecewise function, limits are essential when determining if a function is continuous at the transition between different pieces.
For continuity at a point, the left-hand limit (approaching from the left) and the right-hand limit (approaching from the right) must be equal to the function's value at that point.
  • In the exercise, we calculated \(\lim_{x \to 0^-} g(x) = b\) when approaching from the left with the function \(x + b\).
  • The right-hand limit \((\lim_{x \to 0^+} g(x) = \cos(0) = 1)\) was calculated using \(\cos x\) for \(x \geq 0\).
To find a value of \(b\) that makes the function continuous, we equate the left and right hand limits: \(b = 1\). This ensures that the function’s behavior remains consistent as \(x\) approaches 0 from both directions.
Derivatives
Derivatives represent the rate of change of a function. They are crucial in determining how smooth a function is at a given point, particularly for piecewise functions. A function is said to be differentiable at a point if its derivative exists at that point and matches from both directions.
For the piecewise function discussed:
  • On the left side \(x < 0\), the function \(g(x) = x + b\) has a derivative \(\frac{d}{dx}(x + b) = 1\).
  • While on the \(x \geq 0\) side, the function \(g(x) = \cos x\) has a derivative \(\frac{d}{dx}(\cos x) = -\sin x\), which evaluates to 0 at \(x = 0\).
To check for differentiability, these derivatives must be equal when evaluated at \(x = 0\). Since the left-hand derivative (1) and right-hand derivative (0) do not match, the function is not differentiable at \(x = 0\). This means there is a sharp point or corner at this transition.
Calculus
Calculus encompasses the study of limits, derivatives, integrals, and infinite series, and is a core mathematical discipline employed in a variety of applications from engineering to economics. In our specific exercise, we utilize two essential concepts of calculus: limits and derivatives.
These concepts allow us to analyze the behavior and properties of piecewise functions around troublesome points, such as discontinuities or points of non-differentiability.
  • Limits help us determine whether a piecewise function is continuous at a transition point by comparing the behavior of the function on both sides of the point.
  • Derivatives provide insight into the smoothness and slope of the function curve, indicating how well-connected the transitions in the function are.
Together, these tools of calculus help us design and analyze mathematical models that reflect real-world phenomena. Understanding how functions behave at every point, especially tricky ones, is an integral component of problem-solving in calculus.

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Most popular questions from this chapter

Find the derivative of \(y\) with respect to the given independent variable. $$y=3^{-x}$$

In Exercises \(69-74,\) 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 that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1},\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

a. Let \(Q(x)=b_{0}+b_{1}(x-a)+b_{2}(x-a)^{2}\) be a quadratic approximation to \(f(x)\) at \(x=a\) with these properties: i. \(Q(a)=f(a)\) ii. \(Q^{\prime}(a)=f^{\prime}(a)\) iii. \(Q^{\gamma}(a)=f^{\prime \prime}(a)\) Determine the coefficients \(b_{0}, b_{1},\) and \(b_{2}\) b. Find the quadratic approximation to \(f(x)=1 /(1-x)\) at \(x=0\) c. Graph \(f(x)=1 /(1-x)\) and its quadratic approximation at \(x=0 .\) Then zoom in on the two graphs at the point (0,1) Comment on what you see. d. Find the quadratic approximation to \(g(x)=1 / x\) at \(x=1\) Graph \(g\) and its quadratic approximation together. Comment on what you see. e. Find the quadratic approximation to \(h(x)=\sqrt{1+x}\) at \(x=0 .\) Graph \(h\) and its quadratic approximation together. Comment on what you see. I. What are the linearizations of \(f, g,\) and \(h\) at the respective points in parts (b), (d), and (e)?

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=\frac{3 x+2}{2 x-11}, \quad-2 \leq x \leq 2, \quad x_{0}=1 / 2$$

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=\frac{x^{3}}{x^{2}+1}, \quad-1 \leq x \leq 1, \quad x_{0}=1 / 2$$

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