1. A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford

2. Looking at the Graph By inspection we know: 1.n =3 (cubic) 2.So there are at most 3 real roots. 3.There are at most two (n – 1) extrema points. 4.D f = and R f = 5. y-intercept is (0, 52) Using the TI-89 calculator with window settings: x:[-5, 15] and y:[-400,110] to graph the function which is known as y1(x) = f(x) = x 3 – 9x 2 – 48 + 52 In our calculator

3. Finding The Zeros Methods: 1.We can use our calculator (TI-89) with graph visible a.Choose F5  2:Zero b.Establish an lower and upper bound for each of the zeros. c.Allow the calculator to determine the x-value. 2.Algebraically we can find the zeros we solve the equation f(x) = 0. 3.We can use our TI-89 calculator to do this with the function define to be the y1 variable. The command is solve( y1(x) = 0, x )

4. The solve function solve(y1(x) = 0, x) The solve function can be found starting with the command Home Screen  F2  1:solve which gives us x = 12.51, 0.94, or – 4.44 (rounded to two decimal places. So we have (12.51, 0), (0.94, 0) and (-4.44, 0) as the zeros or x- intercepts for this function

5. Summary of the Analysis

6. Candidates for Extrema From our discussion in class we discovered that we can find candidates for extrema – maximums or minimums – by finding places where the slope of the tangent is 0. That is by solving f’(x) = 0.

7. Using the TI-89 to find Candidates for Extrema To find the candidates for extrema we look for places where the first derivative is zero or undefined. Once again we can use our TI-89 calculator to do this for use. The command is solve( d(y1(x), x) = 0, x ) which solves the equation f ‘(x) = 0 or 3x 2 – 18x -48 = 0 The calculator yields the solutions x = -2 and x = 8. These are the x- values. To find the corresponding y-values evaluate f(-2) and f(8). Note for this to work properly our function must be defined as y1.

8. The True for finding Candidates for Extrema Whenever we are looking for candidates, we must consider: 1.Places where the f’(x) = 0. 2.Places where f’(x) is undefined. 3. If we are working with an interval, then we must consider the endpoints of the interval.

9. Summary of the Analysis

10. Graph of the First Derivative The first derivative is f ‘(x) = 3x 2 -18x – 48 What does this tell us about the function? On the interval (- ∞, - 2) the function is increasing because f’(x) > 0. On the interval (-2, 8) the function is decreasing because f’(x) < 0 On the interval (8, ∞ 0 the function is increasing because f’(x) > 0. How does this compare to the slopes of the tangent on these intervals?

11. Concavity Concavity - the relation of the curve in regards to the tangent. Is it UP or DOWN?

12. The Second Derivative The second derivative is f’’(x) = 6x – 18. What does this say about the first derivative and the graph of f?

13. How to find candidates for the POI We can find candidates for the Point of Inflection (POI) by solving the equation: f’’(x) = 0 Since f’’(x) = 6x – 18. We have x = 3 as the candidate for POI.

14. The TI-89 Candidates for POI The command Solve( d( d(y1(x),x), x) = 0, x) Will solve the equation f’’(x) = 0 for use if we have our function defined in the variable y1.

15. TI-89 and POI The TI-89’s screen should show this The solution is x = 3; that is (3, -146) is the point of the function.

16. The True for finding Candidates for POI Whenever we are looking for candidates, we must consider: 1.Places where the f’’(x) = 0. 2.Places where f’’(x) is undefined. 3. If we are working with an interval, then we must consider the endpoints of the interval.

17. Summary of the Analysis