Problem 70
Question
For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=2 x+4 $$
Step-by-Step Solution
Verified Answer
The function has no points with a horizontal tangent line.
1Step 1: Understand the Problem
To find the points where the tangent line is horizontal, we need to determine where the derivative of the given function is equal to zero, as a horizontal tangent line implies zero slope.
2Step 2: Differentiate the Function
The first step in finding where the tangent line is horizontal is to find the derivative of the function. For the function given, \( y = 2x + 4 \), the derivative with respect to \( x \) is \( y' = \frac{dy}{dx} = 2 \).
3Step 3: Set the Derivative Equal to Zero
To find the points where the tangent line is horizontal, set the derivative equal to zero: \( 2 = 0 \).
4Step 4: Solve for Values of x
From the equation \( 2 = 0 \), we can see that there are no values of \( x \) that satisfy this equation. This indicates that the function does not have any points where the tangent line is horizontal as the derivative is a constant non-zero number.
Key Concepts
Derivative of Linear FunctionSlope of Tangent LineDifferentiation
Derivative of Linear Function
The concept of a derivative is fundamental in calculus, serving as the basis for understanding change. When dealing with linear functions, the derivative holds particular significance. Linear functions are expressed in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
For the given function \( y = 2x + 4 \), the derivative plays a crucial role. The derivative of a linear function, such as this one, is extremely straightforward to find. It simply involves the rate of change, which is constant across the graph of the function. Essentially, the derivative of a linear function is its slope – the factor that multiplies \( x \) in the function, denoted as \( m \).
In our exercise, the derivative is given by \( \frac{dy}{dx} = 2 \). This means that every point on the graph of this function changes at a constant rate of 2, ensuring there are no horizontal tangent lines.
For the given function \( y = 2x + 4 \), the derivative plays a crucial role. The derivative of a linear function, such as this one, is extremely straightforward to find. It simply involves the rate of change, which is constant across the graph of the function. Essentially, the derivative of a linear function is its slope – the factor that multiplies \( x \) in the function, denoted as \( m \).
In our exercise, the derivative is given by \( \frac{dy}{dx} = 2 \). This means that every point on the graph of this function changes at a constant rate of 2, ensuring there are no horizontal tangent lines.
Slope of Tangent Line
Understanding the slope of a tangent line involves grasping how the line touches and complements the curve of a function at a given point. A tangent line can be thought of as a line that gently touches a curve at one specific point, sharing the same gradient or slope as the curve itself at that exact point.
The slope of a tangent line, represented by the derivative of the function, shows how steep the function is at that point. For linear functions like \( y = 2x + 4 \), the slope does not change because the function itself is not curved. The derivative, \( 2 \), indicates this constant slope across all points on the line.
When we talk about horizontal tangent lines, we mean lines where the slope is zero. However, for our function, since the derivative (slope) is always 2, there can never be a point where the slope is zero, which confirms the absence of any horizontal tangents.
The slope of a tangent line, represented by the derivative of the function, shows how steep the function is at that point. For linear functions like \( y = 2x + 4 \), the slope does not change because the function itself is not curved. The derivative, \( 2 \), indicates this constant slope across all points on the line.
When we talk about horizontal tangent lines, we mean lines where the slope is zero. However, for our function, since the derivative (slope) is always 2, there can never be a point where the slope is zero, which confirms the absence of any horizontal tangents.
Differentiation
Differentiation is the process of finding a derivative, and it's the key tool for analyzing how functions change. This process involves using various rules and operations to compute the derivative of a function. In the simplest terms, differentiation is about finding out how a function changes or varies.
In our example function \( y=2x+4 \), differentiation is straightforward because the function is linear. The procedure involves identifying the coefficient of \( x \), as it represents the slope of the function. This slope is 2, derived through an elementary derivative calculation.
The process of differentiation allows us to explore and understand function behaviors, such as where a function reaches its minimum or maximum values, where it is increasing or decreasing, or where horizontal tangent lines might exist. However, given that our function has a constant derivative of 2, it remains straight, leading to the conclusion that no horizontal tangent lines exist.
In our example function \( y=2x+4 \), differentiation is straightforward because the function is linear. The procedure involves identifying the coefficient of \( x \), as it represents the slope of the function. This slope is 2, derived through an elementary derivative calculation.
The process of differentiation allows us to explore and understand function behaviors, such as where a function reaches its minimum or maximum values, where it is increasing or decreasing, or where horizontal tangent lines might exist. However, given that our function has a constant derivative of 2, it remains straight, leading to the conclusion that no horizontal tangent lines exist.
Other exercises in this chapter
Problem 69
Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$ f(x)=\sqrt{x^{2}+\sqrt{1-3 x}} $$
View solution Problem 69
Is the function given by \(G(x)=\frac{1}{x-1}\) continuous over the interval \((0, \infty) ?\) Why or why not?
View solution Problem 70
On the moon, all free-fall distance functions are of the form \(s(t)=0.81 t^{2},\) where \(t\) is in seconds and \(s(t)\) is in meters. An object is dropped fro
View solution Problem 70
Let \(f(x)=\frac{x}{x+1}\) and \(g(x)=\frac{-1}{x+1}\) a) Compute \(f^{\prime}(x)\). b) Compute \(g^{\prime}(x)\). c) What can you conclude about \(f\) and \(g\
View solution