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Section 12.3. The Tangent Line Problem. Tangent Line to a Graph. In Algebra I you learned that the slope of a line indicated the rate at which it rises or falls. For a line this rate (or slope) is the same at every point.
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Section 12.3 The Tangent Line Problem
In Algebra I you learned that the slope of a line indicated the rate at which it rises or falls. • For a line this rate (or slope) is the same at every point. • For graphs other than lines, the rate at which the graph rises or falls changes from point to point.
To determine the rate at which a graph rises or falls at a single point, you find the slope of the tangent line at that point. • In simple terms, the tangent line to the graph of a function at a point is the line that best approximates the slope of the graph at the point.
In geometry, you learned that a line tangent to a circle intersects the circle in exactly one point. • Tangent lines of noncircular graphs can intersect the graph at more than one point.
Because a tangent line approximates the slope of the graph at a point, the problem of finding the slope of a graph at a point is the same as finding the slope of the tangent line at the point.
Example 1 • Use the figure on the next slide to approximate the slope of the graph of f(x) = x3 at the point (1, 1)
In example 1 we found the slope of a graph at a particular point by “eyeballing” the tangent line at that point. • A more precise method of approximating tangent lines makes use of a secant line through the point of tangency and a second point on the graph.
The slope of the secant line through two points is given by • the right side of this equation is called the difference quotient.
The beauty of this procedure is that you obtain a better approximation of the slope of the tangent line by choosing two points closer and closer to the point of tangency.
The slope m of the graph of f at the point (x, f(x)) is equal to the slope of its tangent line at (x, f(x)), and is given by • This definition is a major concept in calculus.
Example 2 • Find the slope of the graph of • f(x) = x3 at the point (2, 8).
Example 3 • Find the slope of f(x) = -3x + 5 using the difference quotient.
Let’s look at the difference in Example 2 and 3. • In Example 2, you were finding the slope of a graph at a specific point (c, f(c)). slope at a specific point
In Example 3, you were finding a formula for the slope at any point on the graph. In such cases, you should use x, rather than c, in the difference quotient. • In all nonlinear graphs this will produce a function of x, which can then be evaluated to find the slope at any desired point. formula for slope
Example 4 • (a)Find a formula for the slope of the graph of f(x) = x2 – 2. • (b)Then find the slopes at the points (-3, 7) and (1, -1).
b. Now find the slope at (-3, 7) and (1, -1). • The slope at (-3, 7) is -6. • The slope at (1, -1) is 2. • End of 1st Day’s Notes
In the last example from yesterday you started with f(x) = x2 – 2 and used the limit process to derive another function m = 2x, that represents the slope of the graph of f at the point (x, f(x)). • This derived function is called the derivative of f at x. It is denoted by • f’(x).
The derivative of f at x is given by • provided this limit exists.
Example 5 • Find the derivative of • f(x) = 4x2 – 5x
There are other notations for a derivative that you will see in calculus. They are