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Chapter 2: Problem 79

Use the alternative form of the derivative to find the derivative at \(x=c\) (if it exists). \(h(x)=|x+5|, \quad c=-5\)

Short Answer

Expert verified
The derivative of the function \(h(x)=|x+5|\) at the point \(x=c=-5\) does not exist.

Step by step solution

01

Find the function's expressions for left and right of x=c

The absolute value function \(|x+5|\) can be expressed differently for the two sides of the point \(x=c=-5\). For \(x<-5\), the function is \(h(x)=-x-5\), and for \(x>-5\), the function is \(h(x)=x+5\).
02

Compute the derivative for the left and right

For \(x<-5\) and for \(x>-5\), the derivative of \(h(x)\) can be computed using the general derivative rule \(h'(x) = \lim_{\Delta x \to 0} \frac{h(x + \Delta x) - h(x)}{\Delta x}\). Here, for \(x<-5\), we get \(h'_{left} = \lim_{\Delta x \to 0} \frac{-x-\Delta x-5-(-x-5)}{\Delta x} = -1\). Similarly, for \(x>-5\), we get \(h'_{right} = \lim_{\Delta x \to 0} \frac{x+\Delta x+5-(x+5)}{\Delta x} = 1\).
03

Check equality of the values at x=c

The derivative will exist at \(x=c=-5\) only if the derivative from the left and right are both the same at that point, that is, if \(h'_{left}(c)= h'_{right}(c)\). However, we have \(h'_{left} = -1\) and \(h'_{right} = 1\). These are not equal, implying that the derivative does not exist at \(x=c=-5\).

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Key Concepts

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

Derivative
In calculus, the derivative of a function is a measure of how a function's output value changes as its input changes. You can think of the derivative as the slope of the tangent line to a curve at a given point. This idea helps to understand how quantities change and enables solving complex real-world problems.

To find the derivative, we analyze the function in respect to the change in its variables. The formal definition involves the limit of the difference quotient:
  • Derivatives describe rates of change.
  • Symbolically, if \( y = f(x) \), then the derivative is \( f'(x) \).
  • The derivative is foundational for applications in physics, economics, and beyond.
Understanding derivatives helps in optimizing processes and predicting future behaviors.
Absolute Value Function
The absolute value function is unique as it always provides the non-negative value of any number or expression. For example, the absolute value of both \( -5 \) and \( 5 \) is \( 5 \). This characteristic makes it very special and interesting to calculate its derivative.

Let's see how it is expressed mathematically:
  • The absolute value function \( |x| \) is defined piecewise.
  • For \( x \geq 0 \), \( |x| = x \).
  • For \( x < 0 \), \( |x| = -x \).
When you locate \( x = -5 \) in \( h(x) = |x+5| \), the distinction between left and right sides of zero must be considered in evaluating the derivative.
Limit Process
The limit process is a fundamental concept in calculus, allowing us to understand the behavior of functions as input values approach a specific point. When finding the derivative, we use this process to measure the instantaneous rate of change at a point by considering what happens as we get infinitely close to that point.

The steps to achieve this are:
  • Apply the classic limit formula \( \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} \).
  • Assess whether the result converges to a single value from both directions.
  • Recognize that if it doesn't, the function has no derivative at that point.
This method gives a precise way to describe how a function behaves exactly at a particular point.
Existence of Derivative
The existence of a derivative at a certain point hinges on the agreement of the derivatives from both sides of that point. In simpler terms, a function has a derivative at a point if the rate of change does not jump or break, no matter the direction from which we approach that point.

Key factors include:
  • Checking if left-hand and right-hand limits match.
  • No sharp corners or cusps at the point in question.
  • In absolute terms, the left and right derivatives must equate.
For \( h(x) = |x+5| \) at \( x = -5 \), the derivatives from either side of the point do not match (\(-1\) and \(1\)), indicating a break in the slope, and thus, no derivative exists at that point.

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