Chapter 3: Problem 3
Find the tangent line approximation to \(1 / x\) near \(x=1\)
Short Answer
Expert verified
The tangent line approximation is \( y = -x + 2 \).
Step by step solution
01
Identify the function and point of approximation
The function to be approximated is \( f(x) = \frac{1}{x} \). We are given the point \( x = 1 \) to find the tangent line approximation.
02
Find the derivative of the function
To find the tangent line, we need the derivative of \( f(x) = \frac{1}{x} \). Using the power rule for derivatives, we get:\[ f'(x) = -\frac{1}{x^2} \]
03
Calculate the derivative at the point x=1
Substitute \( x = 1 \) into the derivative to find the slope of the tangent line:\[ f'(1) = -\frac{1}{1^2} = -1 \]
04
Calculate the value of the function at x=1
Evaluate the original function at \( x = 1 \):\[ f(1) = \frac{1}{1} = 1 \]
05
Write the equation of the tangent line
Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (1, 1) \) and \( m = -1 \):\[ y - 1 = -1(x - 1) \]Simplifying gives:\[ y = -x + 2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
The derivative of a function measures how the function value changes as its input changes. Essentially, it's the slope of the tangent line at any point on the curve of the function. For a function like \( f(x) = \frac{1}{x} \), the derivative tells us how steeply the function rises or falls at each point.
To find the derivative of \( f(x) = \frac{1}{x} \), we use the rule that helps us transform the function into a more manageable form. This form is \( f(x) = x^{-1} \). Then, by applying the power rule for derivatives, we derive that:
\[ f'(x) = -x^{-2} = -\frac{1}{x^2} \]
This tells us that the slope of the tangent to the graph at any point \( x \) is \(-\frac{1}{x^2}\). The derivative is crucial for finding the tangent line, as it provides the necessary slope.
To find the derivative of \( f(x) = \frac{1}{x} \), we use the rule that helps us transform the function into a more manageable form. This form is \( f(x) = x^{-1} \). Then, by applying the power rule for derivatives, we derive that:
\[ f'(x) = -x^{-2} = -\frac{1}{x^2} \]
This tells us that the slope of the tangent to the graph at any point \( x \) is \(-\frac{1}{x^2}\). The derivative is crucial for finding the tangent line, as it provides the necessary slope.
Point-Slope Form
The point-slope form is a method for writing the equation of a line when we know a point on the line and the slope. The general formula is:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope.
In the context of a tangent line, once the derivative gives us the slope at a particular point, we can use point-slope form to find the line's equation. From the exercise, at \(x=1\), the point on the tangent line is \((1, 1)\) since \(f(1)=1\), and the slope is \(m=-1\) as found earlier. Using these values:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope.
In the context of a tangent line, once the derivative gives us the slope at a particular point, we can use point-slope form to find the line's equation. From the exercise, at \(x=1\), the point on the tangent line is \((1, 1)\) since \(f(1)=1\), and the slope is \(m=-1\) as found earlier. Using these values:
- Substitute into point-slope form: \(y - 1 = -1(x - 1)\).
- To simplify, distribute \(-1\): \(y = -x + 1 + 1\).
- The tangent line to the graph is \(y = -x + 2\).
Function Evaluation
Function evaluation involves computing the value of a function for a given input, often to understand the behavior of the function. In algebraic terms, it's simply substituting the input value into the function's formula.
For the function \(f(x) = \frac{1}{x}\), evaluating at \(x = 1\) means calculating:
\[ f(1) = \frac{1}{1} = 1 \]
In this exercise, function evaluation gave us the exact point on the curve where the tangent line needed to touch, ensuring our tangent approximation has the correct point \((x_1, y_1)\) to use in point-slope form.
This step provides the precise vertical coordinate for the approximation, denoting the y-value of the specific point on the curve.
For the function \(f(x) = \frac{1}{x}\), evaluating at \(x = 1\) means calculating:
\[ f(1) = \frac{1}{1} = 1 \]
In this exercise, function evaluation gave us the exact point on the curve where the tangent line needed to touch, ensuring our tangent approximation has the correct point \((x_1, y_1)\) to use in point-slope form.
This step provides the precise vertical coordinate for the approximation, denoting the y-value of the specific point on the curve.
Power Rule
The power rule is a fundamental tool in calculus used to find derivatives. It's particularly useful for functions expressed as powers of \(x\). The rule states:
"If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\)."
This rule allows us to efficiently find the derivative of polynomials. For our function \(f(x) = \frac{1}{x}\), we can rephrase it using the power rule by writing it as \(x^{-1}\). Applying the power rule:
The power rule simplifies the process of differentiation, enabling us to find how a small change in \(x\) affects \(f(x)\). This derivative is critical for determining the slope of the tangent line, thus facilitating the tangent line approximation.
"If \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\)."
This rule allows us to efficiently find the derivative of polynomials. For our function \(f(x) = \frac{1}{x}\), we can rephrase it using the power rule by writing it as \(x^{-1}\). Applying the power rule:
- Identify the exponent: Here, \(n = -1\).
- Differentiate using the power rule: \(f'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}\).
The power rule simplifies the process of differentiation, enabling us to find how a small change in \(x\) affects \(f(x)\). This derivative is critical for determining the slope of the tangent line, thus facilitating the tangent line approximation.