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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.42b
Chapter 3, Problem 3.42b

Derivatives and tangent lines
b. Determine an equation of the line tangent to the graph of f at the point (a,f(a)) for the given value of a.
f(x) = 1/3x-1; a= 2

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Step 1: Identify the function and the point of tangency. The function given is \( f(x) = \frac{1}{3}x - 1 \) and the point of tangency is \( (a, f(a)) \) where \( a = 2 \).
Step 2: Calculate \( f(a) \) to find the y-coordinate of the point of tangency. Substitute \( a = 2 \) into the function: \( f(2) = \frac{1}{3}(2) - 1 \).
Step 3: Find the derivative of the function \( f(x) \) to determine the slope of the tangent line. The derivative \( f'(x) \) of \( f(x) = \frac{1}{3}x - 1 \) is \( f'(x) = \frac{1}{3} \).
Step 4: Evaluate the derivative at \( x = a \) to find the slope of the tangent line at the point. Since \( f'(x) = \frac{1}{3} \), the slope \( m \) at \( x = 2 \) is \( m = \frac{1}{3} \).
Step 5: Use the point-slope form of a line to write the equation of the tangent line. The point-slope form is \( y - f(a) = m(x - a) \). Substitute \( m = \frac{1}{3} \), \( a = 2 \), and \( f(a) \) from Step 2 into the equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivatives

A derivative represents the rate of change of a function at a particular point. It is defined as the limit of the average rate of change of the function as the interval approaches zero. In practical terms, the derivative at a point gives the slope of the tangent line to the graph of the function at that point.
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Derivatives

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. The equation of the tangent line can be expressed in point-slope form, which utilizes the slope and the coordinates of the point of tangency.
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Point-Slope Form

The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. This form is particularly useful for writing the equation of a tangent line once the slope (derivative) and the point of tangency are known. It allows for a straightforward way to express the line based on its slope and a specific point.
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Related Practice
Textbook Question

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1
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