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91Ó°ÊÓ

A function is defined and a closed in terval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of \(c ;\) if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ f(x)=x+|x| ;[-2,1] $$

Short Answer

Expert verified
The Mean Value Theorem does not apply as the function is not differentiable at \(x = 0\).

Step by step solution

01

Review the Mean Value Theorem (MVT)

The Mean Value Theorem states that for any function \(f(x)\) which is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), there exists at least one number \(c\) in the interval \((a, b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b - a}\). This means we first need to check if the function is both continuous and differentiable.
02

Check Continuity

The function \(f(x) = x + |x|\) is a piecewise function:- For \(x \geq 0\), \(|x| = x\) so \(f(x) = 2x\).- For \(x < 0\), \(|x| = -x\) so \(f(x) = 0\).Since the function joins at \(x = 0\) without any break or jump, it is continuous on the interval \([-2, 1]\).
03

Check Differentiability

To check differentiability, we analyze the derivative of \(f(x)\):- For \(x > 0\), \(f(x) = 2x\) so \(f'(x) = 2\).- For \(x < 0\), \(f(x) = 0\) so \(f'(x) = 0\).- At \(x = 0\), the left-hand derivative is 0 and the right-hand derivative is 2, which do not match.Thus, \(f(x)\) is not differentiable at \(x = 0\).
04

Assess MVT Applicability

Since \(f(x)\) is not differentiable on the entire open interval \((-2, 1)\) due to the non-differentiability at \(x = 0\), the Mean Value Theorem does not apply to this function on the interval \([-2, 1]\).
05

Sketch the Graph

To sketch the graph of \(f(x) = x + |x|\) on \([-2, 1]\):- For \(x < 0\), plot the line \(f(x) = 0\), which is the x-axis.- For \(x \geq 0\), plot the line \(f(x) = 2x\), which starts at the origin and moves with a slope of 2.At \(x = 0\), the function connects from the flat line to the slope.

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Key Concepts

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

Continuity
Continuity is an essential property of a function, especially when discussing calculus theorems like the Mean Value Theorem (MVT). A function is continuous on an interval if there are no abrupt jumps, breaks, or holes in its graph over that interval. For the function given in this exercise, \(f(x) = x + |x|\), it is defined piecewise:
  • For \(x \geq 0\), \(f(x) = 2x\).
  • For \(x < 0\), \(f(x) = 0\).
At \(x = 0\), these two parts of the function connect smoothly as there is no sudden change in the function values. Thus, it is continuous across its domain, \([-2, 1]\).
Continuity allows us to predict the function's behavior, ensuring no unexpected disruptions which are critical in applying various mathematical theorems properly.
Differentiability
Differentiability covers whether a function has a derivative, essentially the slope of the tangent line, at every point within an interval. For a function to be differentiable at a given point, its graph around that point should locally resemble a straight line.
For the function \(f(x) = x + |x|\), it is piecewise, and we assess differentiability as follows:
  • For \(x > 0\), the derivative is \(f'(x) = 2\).
  • For \(x < 0\), the derivative is \(f'(x) = 0\).
However, at \(x = 0\), the derivatives from the left side and the right side do not coincide: the left-hand derivative is 0, while the right-hand derivative is 2. This discontinuity in slope causes the function to be non-differentiable at \(x = 0\).
Being non-differentiable at a single point in an interval can impede the applicability of the Mean Value Theorem because differentiability is required across an entire open interval.
Piecewise Function
A piecewise function is a function made up of multiple sub-functions, each applicable to a specific interval. This function, \(f(x) = x + |x|\), is a perfect example of a piecewise function:
  • For \(x \geq 0\), it behaves as \(f(x) = 2x\), a line with slope 2.
  • For \(x < 0\), it is constant, resembling \(f(x) = 0\), laying flat on the x-axis.
When dealing with piecewise functions, it is crucial to pay attention to how the pieces fit together at their boundaries. In this case, at \(x = 0\), both pieces connect smoothly for continuity but diverge in slope, making the function non-differentiable at that point.
Piecewise functions are common in mathematical modeling where a situation can have different behaviors under different circumstances, thus the importance in understanding their continuity and differentiability.

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