A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.
-
Upload
dina-mason -
Category
Documents
-
view
214 -
download
0
Transcript of A discussion of the function f(x) = x 3 – 9x 2 – 48x +52 V. J. Motto University of Hartford.
A discussion of the functionf(x) = x3 – 9x2 – 48x +52
V. J. MottoUniversity of Hartford
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
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 )
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
Summary of the Analysis
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.
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.
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.
Summary of the Analysis
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?
ConcavityConcavity - the relation of the curve in regards to the tangent. Is it UP or DOWN?
The Second Derivative
The second derivative is f’’(x) = 6x – 18. What does this say about the first derivative and the graph of f?
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.
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.
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.
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.
Summary of the Analysis