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 3.4.17

Chapter 3, Problem 3.4.17

Find the slope of the line tangent to the graph of f(x) = x / x+6 at the point (3, 1/3) and at (-2, -1/2).

Verified step by step guidance

Step 1: To find the slope of the tangent line to the graph of a function at a given point, we need to find the derivative of the function, f(x). The function given is f(x) = \(\frac{x}{x+6}\).

Step 2: Use the quotient rule to differentiate f(x). The quotient rule states that if you have a function h(x) = \(\frac{u(x)}{v(x)}\), then h'(x) = \(\frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\). Here, u(x) = x and v(x) = x + 6.

Step 3: Calculate the derivatives u'(x) and v'(x). For u(x) = x, u'(x) = 1. For v(x) = x + 6, v'(x) = 1.

Step 4: Substitute u(x), v(x), u'(x), and v'(x) into the quotient rule formula to find f'(x). This will give you the expression for the derivative of f(x).

Step 5: Evaluate f'(x) at the given points (3, 1/3) and (-2, -1/2) to find the slopes of the tangent lines at these points. Substitute x = 3 and x = -2 into the expression for f'(x) to find the respective slopes.

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Key Concepts

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

Derivative

The derivative of a function at a given point represents the slope of the tangent line to the graph of the function at that point. It is calculated as the limit of the average rate of change of the function as the interval approaches zero. In this context, finding the derivative of f(x) = x / (x + 6) will allow us to determine the slope at specific points.

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. In this problem, we need to find the slope of the tangent line at the points (3, 1/3) and (-2, -1/2) by evaluating the derivative of the function at these x-values.

Point-Slope Form

The point-slope form of a linear equation is used to express the equation of a line given a point on the line and its slope. It is written as y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. This form is useful for constructing the equation of the tangent line once the slope has been determined from the derivative.

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