Problem 13
Question
In Exercises 13 and \(14,\) find the slope of the curve at the indicated point. $$f(x)=|x| \quad\( at \)\quad\( (a) \)x=2 \quad\( (b) \)x=-3$$
Step-by-Step Solution
Verified Answer
The slope of the function \( f(x) = |x| \) at point \( x = 2 \) is \( 1 \) and at point \( x = -3 \) is \( -1 \).
1Step 1: Understanding \( f(x) = |x| \)
The function \( f(x) = |x| \) can be decomposed into two cases: \( f(x) = x \) when \( x \geq 0 \), and \( f(x) = -x \) when \( x < 0 \) because the absolute value of a number is its non-negative value.
2Step 2: Take derivative of \( f(x) = |x| \)
The derivative of the function, \( f'(x) \), will also have two cases: \( f'(x) = 1 \) when \( x > 0 \) and \( f'(x) = -1 \) when \( x < 0 \). The derivative does not exist when \( x = 0 \) as the function is not differentiable at this point.
3Step 3: Find slope at \( x = 2 \)
To find the slope when \( x = 2 \), substitute \( 2 \) into the equation of \( f'(x) \). Because \( 2 > 0 \), the case of \( f'(x) = 1 \) applies here. So, the slope at \( x = 2 \) is \( 1 \).
4Step 4: Find slope at \( x = -3 \)
To find the slope when \( x = -3 \), substitute \( -3 \) into the equation of \( f'(x) \). Here, \( -3 < 0 \), so the case of \( f'(x) = -1 \) applies. Therefore, the slope at \( x = -3 \) is \( -1 \).
Key Concepts
slope of a curvedifferentiabilityabsolute value functions
slope of a curve
When we talk about the slope of a curve at a particular point, we refer to the steepness or incline of the graph at that exact spot. This concept is essentially linked to the derivative of the function at that point.
For a function like the absolute value function, expressed as \(f(x) = |x|\), identifying the slope can be a little tricky due to its piecewise nature.
Let's break it down:
Remember, the slope tells us how sharply the curve rises or falls at any given point!
For a function like the absolute value function, expressed as \(f(x) = |x|\), identifying the slope can be a little tricky due to its piecewise nature.
Let's break it down:
- For \(x > 0\), the function \(f(x) = x\), and hence, the derivative \(f'(x) = 1\). This indicates the slope is 1, meaning the line goes up by 1 unit vertically for every 1 unit it moves horizontally.
- For \(x < 0\), the function becomes \(f(x) = -x\), so \(f'(x) = -1\). Here, the slope is -1, which means the line goes down by 1 unit for each horizontal unit.
Remember, the slope tells us how sharply the curve rises or falls at any given point!
differentiability
Differentiability is a concept that determines if a function has a derivative at a specific point. When dealing with the absolute value function \(f(x) = |x|\), differentiability needs to be scrutinized at \(x = 0\).
Why is \(x = 0\) special? Here are some insights:
Why is \(x = 0\) special? Here are some insights:
- The absolute value function transitions from \(x\) to \(-x\) at \(x = 0\), creating a sharp corner or cusp on the graph. This sudden change in direction prevents us from assigning a single, consistent slope at that point.
- Because of this cusp, the derivative \(f'(x)\) does not exist at \(x = 0\). In mathematical terms, the function is not differentiable here, as it lacks a tangent line that can smoothly touch the graph.
absolute value functions
An absolute value function is written as \(f(x) = |x|\). What makes it unique is its ability to transform both negative and positive inputs into non-negative outputs. This function essentially 'flips' any negative input to its positive counterpart.
Here's how this unfolding works:
This inherently affects its derivative, as seen when calculating the slope at different points. Understanding this nature of absolute value functions helps in analyzing how different inputs affect the output, especially around critical points like zero.
Here's how this unfolding works:
- When \(x \geq 0\), \(f(x) = x\). So, the output is simply the same as the input because the number is already non-negative.
- When \(x < 0\), \(f(x) = -x\). Here, the output is the positive version of the input since multiplying by -1 negates the negative sign.
This inherently affects its derivative, as seen when calculating the slope at different points. Understanding this nature of absolute value functions helps in analyzing how different inputs affect the output, especially around critical points like zero.
Other exercises in this chapter
Problem 12
In Exercises \(9-12,\) at the indicated point find (a) the slope of the curve, (b) an equation of the tangent, and (c) an equation of the tangent. (d) Then draw
View solution Problem 12
In Exercises \(11-18,\) use the function \(f\) defined and graphed below to answer the questions. $$f(x)=\left\\{\begin{array}{ll}{x^{2}-1,} & {-1 \leq x
View solution Problem 13
In Exercises \(7 - 14 ,\) determine the limit by substitution. Support graphically. $$\lim _ { x \rightarrow - 2 } ( x - 6 ) ^ { 2 / 3 }$$
View solution Problem 13
In Exercises \(11-18,\) use the function \(f\) defined and graphed below to answer the questions. $$f(x)=\left\\{\begin{array}{ll}{x^{2}-1,} & {-1 \leq x
View solution