A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.
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Transcript of A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.
![Page 1: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/1.jpg)
A discussion of the functionf(x) = x3 – 9x2 – 48x +52
V. J. MottoUniversity of Hartford
![Page 2: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/2.jpg)
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.Df = and Rf =
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) = x3 – 9x2 – 48 + 52
In our calculator
![Page 3: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/3.jpg)
Finding The ZerosMethods:
1.We can use our calculator (TI-89) with graph visible a. Choose F52:Zerob. 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 )
![Page 4: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/4.jpg)
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
![Page 5: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/5.jpg)
Summary of the Analysis
![Page 6: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/6.jpg)
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.
![Page 7: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/7.jpg)
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 3x2 – 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.
![Page 8: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/8.jpg)
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.
![Page 9: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/9.jpg)
Summary of the Analysis
![Page 10: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/10.jpg)
Graph of the First Derivative
The first derivative is f ‘(x) = 3x2 -18x – 48What 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) < 0On 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?
![Page 11: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/11.jpg)
ConcavityConcavity - the relation of the curve in regards to the tangent. Is it UP or DOWN?
![Page 12: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/12.jpg)
The Second Derivative
The second derivative is f’’(x) = 6x – 18. What does this say about the first derivative and the graph of f?
![Page 13: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/13.jpg)
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.
![Page 14: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/14.jpg)
The TI-89 Candidates for POI
The commandSolve( 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.
![Page 15: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/15.jpg)
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.
![Page 16: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/16.jpg)
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.
![Page 17: A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.](https://reader036.fdocuments.in/reader036/viewer/2022082710/56649e4b5503460f94b3fd28/html5/thumbnails/17.jpg)
Summary of the Analysis