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Chapter 2: Problem 85

Determining Differentiability In Exercises \(85-88\) , find the derivatives from the left and from the right at \(x=1\) (if they exist). Is the function differentiable at \(x=1 ?\) $$f(x)=|x-1|$$

Short Answer

Expert verified
The function \(f(x) = |x - 1|\) is not differentiable at \(x = 1\).

Step by step solution

01

Rewrite the absolute value function as a piecewise function

The absolute value function \(f(x) = |x - 1|\) can be written as a piecewise function as follows: \[f(x) = \begin{cases} x - 1 & \text{if } x \geq 1 \ 1 - x & \text{if } x < 1 \end{cases}\].
02

Find the derivative from the left at \(x = 1\)

To find the derivative of \(f(x)\) from the left at \(x = 1\), consider the function when \(x < 1\), which is \(f(x) = 1 - x\). The derivative of \(f(x)\) with respect to \(x\) would be \(-1\).
03

Find the derivative from the right at \(x = 1\)

The derivative from the right at \(x = 1\) can be found by considering the function when \(x \geq 1\), which is \(f(x) = x - 1\). The derivative of \(f(x)\) with respect to \(x\) would be 1.
04

Check if the function is differentiable at \(x = 1\)

Since the value of the derivative from the left at \(x = 1\) does not equal to the derivative from the right at \(x = 1\), the function is not differentiable at \(x = 1\).

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Key Concepts

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

Derivative
A derivative represents how a function changes as its input changes. For a given function, the derivative at a certain point provides the slope of the tangent line to the function at that point.
In simpler terms:
  • It tells us how "steep" the function is at any specific point.
  • A positive derivative means the function is increasing, while a negative one means it's decreasing.
Calculating a derivative involves taking the limit of the average rate of change of a function over an interval, as the interval approaches zero. This is often expressed as follows:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.\]The above formula defines the derivative at any point \(x\). Understanding derivatives is key to understanding how functions behave locally.
Piecewise Function
A piecewise function is composed of multiple "pieces" or segments, each defined by a specific condition or range of input values.
Piecewise functions are useful for describing situations where a rule or effect changes based on different circumstances.
The piecewise function for the absolute value of \(x - 1\) is:\[ f(x) = \begin{cases} x - 1 & \text{if } x \geq 1 \ 1 - x & \text{if } x < 1 \end{cases}\]Here's what happens:
  • For \(x\) values greater than or equal to 1, the function behaves like \(x - 1\).
  • For \(x\) values less than 1, it behaves like \(1 - x\).
Piecewise functions let us model scenarios where the behavior of a system changes across different conditions.
Absolute value
The absolute value function, denoted by \(|x|\), measures the "magnitude" of a number without considering its sign. It transforms negative numbers into their positive counterparts and leaves positive numbers unchanged.
For a real number \(x\):
  • If \(x \geq 0\), then \(|x| = x\).
  • If \(x < 0\), then \(|x| = -x\).
In the exercise, the function \(f(x) = |x - 1|\) can be rewritten as a piecewise function due to the definition of absolute value. This nature of absolute value having different expressions based on the sign of its argument leads to cases where a function may not be smooth at a certain point.
The absolute value function frequently appears in real-world applications, such as measuring distance."
Left-hand derivative
The left-hand derivative of a function at a given point considers only values of the function that are less than the point.
It is a measure of how the function behaves as it approaches the point from the left side.
To find the left-hand derivative at \(x = 1\) for the function \(f(x) = |x - 1|\), you examine the part of the piecewise function defined for \(x < 1\), which is \(1 - x\).
The derivative of \(1 - x\) is \(-1\).
In this scenario:
  • If moving towards a specific point from the left results in a different slope compared to moving from the right, this typically indicates a discontinuity or corner point in the graph at that specific place.
The left-hand derivative can help identify points where a function's behavior sharply changes.
Right-hand derivative
The right-hand derivative considers the values of a function that are greater than or equal to a given point. It indicates how the function is changing as it approaches the point from the right.
For our example, to determine the right-hand derivative at \(x = 1\), look at the segment where \(x \geq 1\), or \(x - 1\).
The derivative is simply 1.
If we compare the right-hand and left-hand derivatives calculated earlier:
  • The right-hand derivative is 1, while the left-hand derivative is \(-1\).
In order for a function to be considered differentiable at a point, both the left-hand and right-hand derivatives must agree, indicating a smooth transition.
In this case, they do not match, pointing out a lack of differentiability at \(x = 1\).

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Most popular questions from this chapter

Harmonic Motion The displacement from equilibrium of an object in harmonic motion on the end of a spring is \(y=\frac{1}{3} \cos 12 t-\frac{1}{4} \sin 12 t\) where \(y\) is measured in feet and \(t\) is the time in seconds. Determine the position and velocity of the object when \(t=\pi / 8 .\)

pumped into the trough. Moving Ladder A ladder 25 feet long is leaning against the wall of a house (see figure). The base of the ladder is pulled away from the wall at a rate of 2 feet per second. (a) How fast is the top of the ladder moving down the wall when its base is 7 feet, 15 feet, and 24 feet from the wall? (b) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. (c) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall.

Finding a Second Derivative In Exercises \(83-88\) , find the second derivative of the function. $$f(x)=6\left(x^{3}+4\right)^{3}$$

Finding a Second Derivative In Exercises \(83-88\) , find the second derivative of the function. $$f(x)=\sin x^{2}$$

Proof Let \(u\) be a differentiable function of \(x .\) Use the fact that \(|u|=\sqrt{u^{2}}\) to prove that \(\frac{d}{d x}[|u|]=u^{\prime} \frac{u}{|u|}, \quad u \neq 0.\)
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