If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
010. `query 10
Question:
`qWhat is a polynomial with zeros at -3, 4 and 9? Describe the graph of your polynomial.
Your solution:
Confidence Rating:
Given Solution:
`a A polynomial with zeros at -3, 4 and 9 must have factors
(x + 3), ( x - 4) and (x - 9), and so must contain (x+3) (x-4) (x-9) as
factors.
These factors can be multiplied by any constant. For example
8 (x+3) (x-4) (x-9),
-2(x+3) (x-4) (x-9) and
(x+3) (x-4) (x-9) / 1872
are all polynomials with zeros at -3, 4 and 9.
If -3, 4 and 9 are the only zeros then (x+3), (x-4) and (x-9) are the only possible
linear factors.
It is possible that the polynomial also has irreducible quadratic
factors. For example x^2 + 3x + 10 has
no zeros and is hence irreducible, so
(x+3) (x-4) (x-9) (x^3 + 3x + 10) would also be a polynomial
with its only zeros at -3, 4 and 9.
The polynomial could have any number of irreducible
quadratic factors. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q1.5.18 (was 1.5.16 right-, left-hand limits and limit (sin
fn)
Self-critique (if necessary):
Self-critique Rating:
Question:
`qWhat are the three limits for your function (if a limit
doesn't exist say so and tell why)?
Your solution:
Confidence Rating:
Given Solution:
`a Imagine walking along the graph from the right. As you approach the limiting x value, your y
'altitude' gets steadier and steadier, approaching closer and closer what
value? You don't care what the function
actually does at the limiting value of x, just how it behaves as you approach
that limiting value.
The same thing happens if you walk along the graph from the
left. What does you y value approach?
Is this the value that you approach as you 'walk in' from
the right, as well as from the left? If
so then it's your limit. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q1.5.22 (was 1.5.20 right-, left-hand limits and limit
(discont at pt)
What are the three limits for your function (if a limit
doesn't exist say so and tell why)?
Your solution:
Confidence Rating:
Given Solution:
`aSTUDENT RESPONSE:
The three limits are a. 0
b. 2 c. no limit The limit doesn't exist because while the
function approaches left and right-handed limits, those limite are different.
INSTRUCTOR COMMENT:
That is correct.
ADVICE TO ALL STUDENTS:
Remember that the limit of a function at a point depends
only on what happens near that point.
What happens at the point iself is irrelevant to the limiting behavior
of the function as we approach that point.
Would it be ok to say that the two limits are different
because the graph indicates -1 being undefined?
INSTRUCTOR RESPONSE
The limit as x approaches -1 isn't affected by the value of
the function at x = -1, or by the lack of a value.
You consider only what happens as you get close to -1. In considering the limit,
you never actually get to -1.
Self-critique (if necessary):
Self-critique Rating:
Question:
`q1.5.30 (was 1.5.26 lim of (x+4)^(1/3) as x -> 4
What is the desired limit and why?
Your solution:
Confidence Rating:
Given Solution:
The limit of a function at a point is not
defined by its value at that point, but rather by its values
near the point.
A continuous function is one for which the values near a given
point happen to approach the value of the function at that
point. This is the case for most familiar functions at most points, so it can be
difficult to appreciate why this is an important definition.
The present function f(x) = (x + 4)^(1/3) is defined for all values of x, and
like most functions with which you are familiar, is continuous for all values of
x.
The present question asks about the limit of this function as x approaches 4.
If x is near 4, then in this case f(x) is near f(4).
Furthermore no matter how close we want f(x) to be to f(4), we can guarantee that it's this close by requiring that x be close enough to 4.
It is for this reason and in this sense that the limiting value of f(x) = (x + 4)^(1/3), as x approaches 4, is (4 + 4)^(1/3) = 2.
For example, if we want f(x) to be within .001 of 2, this can certainly be guaranteed by letting x be within .00005 of 4. You might well ask how we came up with .0001 (answer: if y = (x+4)^(1/3) then x = y^3 - 4; the cube of a number that lies within delta of 2 is (2 - delta)^3 = 2^3 - 3 delta * 2^2 + 3 delta^2 * 2 - delta^3 = 8 - 12 delta + 6 delta^2 - delta^3; we use this to reason out how close x must be to 4 in order to guarantee that y be within .001 of 2.
This boils down to the following:
The function is continuous, so f(c) = c and in this case the limit is 8^(1/3), which is 2.
Self-critique (if necessary):
Self-critique Rating:
Question:
`q1.5.48 (was 1.5.38 lim of (x^3-1)/(x-1) as x -> 1
Your solution:
Confidence Rating:
Given Solution:
`a As x -> 0 both numerator and denominator approach 0, so you can't tell just by plugging in numbers what the limit will be.
The given function is not defined for x = 0, so it cannot be evaluated at x = 0.
However if you factor x^3-1 into (x-1)(x^2+x+1) you can reduce the fraction to (x-1)(x^2+x+1) / (x-1) = x^2 + x + 1.
This expression is not the same as the original function, because this is defined at x = 0, whereas the original expression was not defined at x = 0.
In any case the expression x^2 + x + 1 is equal to the
original function for all x except 1 (we can't reduce for x = 1 because x-1
would be zero, and we can't divide by zero).
The expression x^2 + x + 1 is continuous, so as x -> 1, x^2 + x + 1 -> 3, and the limit of our original function, as x -> 1, is 3.
It doesn't matter at all what the function does at x = 1, because the limiting value of x is not important when you take the limit--only x values approaching the limit matter.
STUDENT COMMENT
This is a little confusing. I don’t completely understand the
steps that are written out here.
INSTRUCTOR RESPONSE
In order for me to clarify, you need to tell me as
specifically as possible what you do and do not understand.
I expect that you understand how factoring reduces the expression to x^2 + x +
1.
I expect you understand that when x is near 1 the value of x^2 + x + 1 is near
3.
The likely point of confusion is the rest of the explanation, which establishes
why the expression x^2 + x + 1 applies only as long as x is not 1, and why this
doesn't matter when finding the limit, which considers only values of the
function when x is 'near' 1, what happens at x = 1.
Let me know the specifics about what you do and do not understand, and I'll be glad to clarify further.
Self-critique (if necessary):
Self-critique Rating:
Question:
`q1.5.70 (was 1.5.56
lim of 1000(1+r/40)^40 as r -> 6%
Your solution:
Confidence Rating:
Given Solution:
`a $1000 *( 1+.06 / 40)^40 = 1061.788812.
Since this function changes smoothly as you move through r =
.06--i.e., since the function is continuous at r = .06--this value will be the
limit. **
Add comments on any
surprises or insights you experienced as a result of this assignment.
Self-critique (if necessary):
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