|Module 9 - The Relationship between a Function and Its First and Second Derivative|
|Introduction | Lesson 1 | Lesson 2 | Self-Test|
|Lesson 9.1: What the First Derivative Says About the Function|
In Module 8 we saw that the value of the derivative of f at x is given by the slope of the line tangent to the graph of f at x. In this lesson you will explore what the first derivative says about the graph of the original function by using the Derivative and Tangent features of the calculator.
If f is a function, then its first derivative is denoted by f ' , which is read "f prime," and the value of the first derivative at x = a is f '(a). The calculator denotes the first derivative of f as , which is an alternate notation. If y = f(x), other notations include , and dy/dx.
Using the Derivative Feature on the Graph Screen
Follow the procedure given below to graph a function and use the Derivative feature of the Graph screen's Math menu to compute its derivative.
Find the value of the derivative at x = -2 with the Derivative feature in the F5:Math menu.
You should see the graph and a prompt for the x-coordinate at which the derivative is to be evaluated.
The value of the derivative of f(x) = x3 2x2 5x + 6 at x = -2 is 15.
Using the Tangent Feature
You can use the Tangent feature of the Graph Math menu to see the graph of a tangent line at a specific point along with the equation for that tangent line.
After you make the selection, you should see the graph of the function and a prompt for the x-coordinate of the point of tangency.
The TI-89 displays the graph of the tangent line and its equation is shown at the bottom of the screen. The equation of the tangent line in the example above is y = 15x + 30.
9.1.1 How does the equation of the tangent line display the value of the derivative at x = -2? Click here for the answer.
9.1.2 The table below contains points on the graph of the function f(x) = x3 2x2 5x + 6. Use the Derivative feature to find the derivative at each point shown and record the values.
Click here for the answer.
9.1.3 What do you notice about the function values on the intervals where the derivative is positive compared with the function values on the intervals where the derivative is negative?
Finding Turning Points from the Derivative
The function is increasing exactly where the derivative is positive, and decreasing exactly where the derivative is negative. On the graph of the derivative find the x-value of the zero to the left of the origin.
One of the zeros of the derivative is approximately x = -0.7863.
9.1.4 Use the Zero feature of the F5:Math menu to find the other zero of the derivative. Click here for the answer.
Summarizing the Relationship between f and f '
The following characteristics of the function f(x) = x3 2x2 5x + 6 can be determined from the graph of its first derivative.
The derivative is zero at x = -0.7863 and at x = 2.11963, which are the x-values of the turning points.
The function is increasing on the intervals (- , -0.7863) and (2.11963, ) because its first derivative is positive there.
The function is decreasing on the interval (-0.7863, 2.11963) because its first derivative is negative there.
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