Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.4.61a

Chapter 3, Problem 3.4.61a

Find an equation of the line tangent to the given curve at a.
y = (x + 5) / (x - 1); a = 3

Verified step by step guidance

Step 1: Identify the function and the point of tangency. The function is $ y = \frac{x + 5}{x - 1} $ and the point of tangency is at $ x = 3 $.

Step 2: Calculate the derivative of the function to find the slope of the tangent line. Use the quotient rule: if $ y = \frac{u}{v} $, then $ y' = \frac{u'v - uv'}{v^2} $. Here, $ u = x + 5 $ and $ v = x - 1 $.

Step 3: Differentiate $ u $ and $ v $. $ u' = 1 $ and $ v' = 1 $. Substitute these into the quotient rule to find $ y' $.

Step 4: Evaluate the derivative at $ x = 3 $ to find the slope of the tangent line. Substitute $ x = 3 $ into the expression for $ y' $.

Step 5: Use the point-slope form of a line, $ y - y_1 = m(x - x_1) $, where $ m $ is the slope found in Step 4 and $ (x_1, y_1) $ is the point $ (3, y(3)) $. Calculate $ y(3) $ using the original function.

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Key Concepts

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

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 represents the instantaneous rate of change of the function at that point, which can be found using the derivative.

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. For the function y = (x + 5) / (x - 1), the derivative will provide the slope of the tangent line at the point where x = 3.

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 (from the derivative) and the point of tangency (the coordinates of the curve at x = 3) are known.

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Find an equation of the line tangent to the given curve at a.y = (x + 5) / (x - 1); a = 3

Verified video answer for a similar problem:

Key Concepts

Tangent Line

Derivative

Point-Slope Form

Find an equation of the line tangent to the given curve at a.
y = (x + 5) / (x - 1); a = 3