course Mth 271 010. `query 10*********************************************
.............................................
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 polynomizls 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 Rating: I used logic to solve this problem. ********************************************* Question: `qWhat are the three limits for your function (if a limit doesn't exist say so and tell why)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This problem doesn’t have any limits. Limits exist when a problem is divided by x and some number. A limit would be the number that would make the denominator 0. For this problem, the function isn’t divided by anything, so there are no limits. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I feel fairly confident about this problem.
.............................................
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 limits 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 itself is irrelevant to the limiting behavior of the function as we approach that point. ------------------------------------------------ Self-critique Rating: I think I understand the problem. The given solution makes no sense to me.
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..............................................
Given Solution: `a The function is continuous, so f(c) = c and in this case the limit is 8^(1/3), which is 2. ** ------------------------------------------------ Self-critique Rating: I plugged the value of the limit into the equation and solved. ********************************************* Question: `q lim of (x^3-1)/(x-1) as x -> 1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (x^3-1)/(x-1) as x -> 1 (x – 1)(x^2 + x – 1)/(x – 1) x^2 + x + 1 (1)^2 + (1) + 1 = 3 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I feel very confident about this problem.
.............................................
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. 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, which 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). As x -> 1, x^2 + x + 1 -> 3. It doesn't matter at all what the function does at x = 1, because the limiting value of x never occurs when you take the limit--only x values approaching the limit count. 3 is therefore the correct limit. **
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 style='mso-spacerun:yes'> 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. style='mso-spacerun:yes'> ------------------------------------------------ Self-critique Rating: For this problem, I expanded the numerator so I could eliminate the denominator and plugged the limit value in for x. ********************************************* Question: `q1.5.70 (was 1.5.56 lim of 1000(1+r/40)^40 as r -> 6% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1000(1 + r/40)^40 r -> .06 1000(1 + (.06)/40)^40 = 1061.79 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I feel very confident about this problem..............................................
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 Rating: I used the same procedure as previous problems to solve this. "