Problem 50
Question
Use a graphing utility to make rough estimates of the locations of all horizontal tangent lines, and then find their exact locations by differentiating. $$y=\frac{x^{2}+9}{x}$$
Step-by-Step Solution
Verified Answer
Horizontal tangents are at points (3, 6) and (-3, -6).
1Step 1: Rewrite the Function
Rewrite the function in a form suitable for differentiation. Given the function \( y = \frac{x^2 + 9}{x} \), simplify it to \( y = x + \frac{9}{x} \).
2Step 2: Differentiate the Function
Find the derivative of the function \( y = x + \frac{9}{x} \). Use the power rule for differentiation: \( y' = 1 - \frac{9}{x^2} \).
3Step 3: Determine Horizontal Tangents
A horizontal tangent occurs where the derivative is zero. Set the derivative equal to zero and solve: \( 1 - \frac{9}{x^2} = 0 \).
4Step 4: Solve for x-values
Solve the equation \( 1 = \frac{9}{x^2} \). Multiply both sides by \( x^2 \) to get \( x^2 = 9 \).
5Step 5: Find Exact Locations
Solve \( x^2 = 9 \) to find \( x \) values by taking the square root: \( x = 3 \) or \( x = -3 \). Thus, horizontal tangents occur at \( x = 3 \) and \( x = -3 \).
6Step 6: Verify Points on Original Function
Substitute \( x = 3 \) and \( x = -3 \) into the original function to find the corresponding \( y \) values. For \( x = 3 \), \( y = \frac{3^2 + 9}{3} = 6 \). For \( x = -3 \), \( y = \frac{(-3)^2 + 9}{-3} = -6 \).
7Step 7: Provide Exact Points
The points where horizontal tangents occur are \( (3, 6) \) and \( (-3, -6) \).
Key Concepts
DifferentiationHorizontal TangentsGraphing UtilityPower Rule
Differentiation
Differentiation is a fundamental concept in calculus, focusing on how functions change. In simple terms, differentiation allows us to find the rate at which one quantity changes with respect to another. It is the process of finding a derivative, which essentially gives you an equation that can tell you the slope of the tangent line to any point on a curve.
To differentiate a function, we often rewrite it to make the process easier. For example, if you have a fraction like \( y = \frac{x^2 + 9}{x} \), the first step is to simplify it to \( y = x + \frac{9}{x} \). This makes the function easier to handle for differentiation using the Power Rule. By differentiating, you find the changes in the function's output as its input changes, which is crucial for understanding the behavior of curves.
To differentiate a function, we often rewrite it to make the process easier. For example, if you have a fraction like \( y = \frac{x^2 + 9}{x} \), the first step is to simplify it to \( y = x + \frac{9}{x} \). This makes the function easier to handle for differentiation using the Power Rule. By differentiating, you find the changes in the function's output as its input changes, which is crucial for understanding the behavior of curves.
Horizontal Tangents
Horizontal tangents are the points on a curve where the slope of the tangent line is zero. In other words, at these points, the curve becomes flat. Finding horizontal tangents is essential for understanding areas where a function switches direction or reaches a local maximum or minimum.
To locate horizontal tangents for the function \( y = x + \frac{9}{x} \), we first differentiate to find \( y' \). We discovered earlier that \( y' = 1 - \frac{9}{x^2} \). Next, we set \( y' \) equal to zero because this is where the slope is horizontal. Solving \( 1 - \frac{9}{x^2} = 0 \), you find \( x = 3 \) and \( x = -3 \). These are the points where the function's graph flattens out, meaning the tangent is horizontal.
To locate horizontal tangents for the function \( y = x + \frac{9}{x} \), we first differentiate to find \( y' \). We discovered earlier that \( y' = 1 - \frac{9}{x^2} \). Next, we set \( y' \) equal to zero because this is where the slope is horizontal. Solving \( 1 - \frac{9}{x^2} = 0 \), you find \( x = 3 \) and \( x = -3 \). These are the points where the function's graph flattens out, meaning the tangent is horizontal.
Graphing Utility
A graphing utility is a valuable tool for estimating and visualizing the behavior of mathematical functions. This technology can plot functions quickly, making it easier to spot features like intercepts, turning points, and tangents.
For our particular function \( y = \frac{x^2 + 9}{x} \), using a graphing utility allows you to approximate where horizontal tangents might occur before calculating them exactly. You can visually confirm by plotting the function that at \( x = 3 \) and \( x = -3 \), the tangent lines look horizontal. However, this doesn't replace calculus—it's a method to get a sneak peek at what to compute, helping to ensure accuracy when proceeding with mathematical operations.
For our particular function \( y = \frac{x^2 + 9}{x} \), using a graphing utility allows you to approximate where horizontal tangents might occur before calculating them exactly. You can visually confirm by plotting the function that at \( x = 3 \) and \( x = -3 \), the tangent lines look horizontal. However, this doesn't replace calculus—it's a method to get a sneak peek at what to compute, helping to ensure accuracy when proceeding with mathematical operations.
Power Rule
The Power Rule is a quick and straightforward method of differentiation used in calculus to find the derivative of functions involving variables raised to a power. If you have a function of the form \( x^n \), where \( n \) is any real number, the Power Rule states that the derivative is \( nx^{n-1} \).
Consider the differentiated expression \( y = x + \frac{9}{x} \) from the exercise. We break it down into individual parts: \( x^1 \) and \( 9x^{-1} \). Using the Power Rule, the derivative of \( x^1 \) is \( 1 \times x^{1-1} = 1 \), and the derivative of \( 9x^{-1} \) is \( 9 \times (-1)x^{-2} = -\frac{9}{x^2} \). This leads us to the derivative \( y' = 1 - \frac{9}{x^2} \). The Power Rule simplifies the process of finding derivatives, transforming complex expressions into more manageable ones.
Consider the differentiated expression \( y = x + \frac{9}{x} \) from the exercise. We break it down into individual parts: \( x^1 \) and \( 9x^{-1} \). Using the Power Rule, the derivative of \( x^1 \) is \( 1 \times x^{1-1} = 1 \), and the derivative of \( 9x^{-1} \) is \( 9 \times (-1)x^{-2} = -\frac{9}{x^2} \). This leads us to the derivative \( y' = 1 - \frac{9}{x^2} \). The Power Rule simplifies the process of finding derivatives, transforming complex expressions into more manageable ones.
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