Problem 22
Question
Calculate \(f^{\prime}(x),\) and sketch the graph of the equation \(y=f^{\prime}(x)\). $$ f(x)=3 x-|x| $$
Step-by-Step Solution
Verified Answer
The derivative is piecewise: \( f'(x) = 2 \) if \( x \geq 0 \) and \( f'(x) = 4 \) if \( x < 0 \). The graph has two horizontal lines at \( y=2 \) and \( y=4 \).
1Step 1: Identify the Function Components
The function given is \( f(x) = 3x - |x| \). This function is composed of two components: \( 3x \), a linear function, and \( |x| \), an absolute value function. We need to consider how these components behave for different values of \( x \).
2Step 2: Understand the Absolute Value Function
The absolute value function \( |x| \) behaves differently depending on whether \( x \) is positive or negative. \( |x| = x \) when \( x \geq 0 \) and \( |x| = -x \) when \( x < 0 \). This distinction will help us express the function \( f(x) \) without the absolute value.
3Step 3: Express the Function in Piecewise Form
Based on the behavior of \( |x| \), we can express \( f(x) \) as a piecewise function:- If \( x \geq 0 \): \( f(x) = 3x - x = 2x \).- If \( x < 0 \): \( f(x) = 3x - (-x) = 4x \).
4Step 4: Determine the Derivative for Each Case
Now, find the derivative \( f'(x) \) for each piece of the piecewise function:- If \( x \geq 0 \): \( f(x) = 2x \) implies \( f'(x) = 2 \).- If \( x < 0 \): \( f(x) = 4x \) implies \( f'(x) = 4 \).
5Step 5: Construct the Derivative Function
Combine the derivatives into a piecewise function:- \( f'(x) = \begin{cases} 2 & \text{if } x \geq 0 \ 4 & \text{if } x < 0 \end{cases} \).
6Step 6: Sketch the Graph of the Derivative Function
To graph \( y = f'(x) \), plot two constant lines:- A horizontal line at \( y = 2 \) for all \( x \geq 0 \).- A horizontal line at \( y = 4 \) for all \( x < 0 \). These two lines represent the piecewise nature of \( f'(x) \) and indicate a step change at \( x = 0 \).
Key Concepts
Piecewise FunctionsDerivativesAbsolute Value Functions
Piecewise Functions
Piecewise functions are mathematical expressions that have different forms for different parts of their domain. They allow us to construct functions that can change behavior at specific points. In our exercise, we have a function \( f(x) = 3x - |x| \), which can be expressed as different linear expressions based on the value of \( x \).
- When \( x \geq 0 \), \( f(x) \) simplifies to \( 2x \).- When \( x < 0 \), \( f(x) \) becomes \( 4x \).
This kind of breakdown showcases a core characteristic of piecewise functions: they manage different conditions or intervals separately but within the same overall function. This piecewise nature makes it easier to calculate derivatives, as each piece is a standard function, either a line or a polynomial, allowing for straightforward differentiation.
- When \( x \geq 0 \), \( f(x) \) simplifies to \( 2x \).- When \( x < 0 \), \( f(x) \) becomes \( 4x \).
This kind of breakdown showcases a core characteristic of piecewise functions: they manage different conditions or intervals separately but within the same overall function. This piecewise nature makes it easier to calculate derivatives, as each piece is a standard function, either a line or a polynomial, allowing for straightforward differentiation.
Derivatives
In calculus, derivatives represent the rate at which a function changes. They are crucial for understanding the behavior of functions. For piecewise functions like our example \( f(x) = 3x - |x| \), derivatives can be determined separately for each piece.
Thus, the derivative function \( f'(x) \) itself is a piecewise function:\[ f'(x) = \begin{cases} 2 & \text{if}\ x \geq 0 \ 4 & \text{if}\ x < 0 \end{cases} \]
Knowing the derivative allows us not only to graph the function's rate of change but also to anticipate how the function values develop over time and different intervals.
- For \( f(x) = 2x \) when \( x \geq 0 \), the derivative \( f'(x) \) is simply \( 2 \), representing a constant rate of change.
- For \( f(x) = 4x \) when \( x < 0 \), the derivative \( f'(x) \) is \( 4 \), again a constant but at a steeper slope.
Thus, the derivative function \( f'(x) \) itself is a piecewise function:\[ f'(x) = \begin{cases} 2 & \text{if}\ x \geq 0 \ 4 & \text{if}\ x < 0 \end{cases} \]
Knowing the derivative allows us not only to graph the function's rate of change but also to anticipate how the function values develop over time and different intervals.
Absolute Value Functions
The absolute value function, denoted as \(|x|\), is fundamental in calculus as it measures the "distance" from zero, ignoring direction. It transforms values to non-negative equivalents, like converting \(-5\) to \(5\).
- When \( x \geq 0 \), \(|x| = x\).- When \( x < 0 \), \(|x| = -x\).
This duality captures the essence of absolute value: treating all numbers, whether positive or negative, in terms of their magnitude only. This characteristic allows \(|x|\) to effectively alter expressions and equations, such as in our function \( f(x) = 3x - |x| \), transforming a linear equation into a versatile piecewise function. Absolute values contribute significantly to piecewise functions because they introduce necessary conditions based on the behavior of the input values, distinguishing how the function reacts to different parts of its domain.
- When \( x \geq 0 \), \(|x| = x\).- When \( x < 0 \), \(|x| = -x\).
This duality captures the essence of absolute value: treating all numbers, whether positive or negative, in terms of their magnitude only. This characteristic allows \(|x|\) to effectively alter expressions and equations, such as in our function \( f(x) = 3x - |x| \), transforming a linear equation into a versatile piecewise function. Absolute values contribute significantly to piecewise functions because they introduce necessary conditions based on the behavior of the input values, distinguishing how the function reacts to different parts of its domain.
Other exercises in this chapter
Problem 22
Use the Reciprocal Rule to compute the derivative of the given expression with respect to \(x\) $$ 3 /(2+\sin (x)) $$
View solution Problem 22
Find the slope of the tangent line to the graph of the given function at the given point \(P\). $$ f(x)=x^{3}-2 x \quad P=(1,-1) $$
View solution Problem 23
Differentiate the given expression with respect to \(x\). $$ \log _{2}\left(\arccos \left(x^{2}\right)\right) $$
View solution Problem 23
Use implicit differentiation to find the tangent line to the given curve at the given point \(P_{0}\). \(\ln (x y-1)+y^{2}=4 \quad P_{0}=(1,2)\)
View solution