#$&*
course Mth 271
11/9 730p
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Question: `q 1 c 7th edition 2.4.12 (was 2.4.10) der of f(x) = (x+1)/(x-1) at (2,3)
What is the derivative of f(x) at the given point?
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Your solution:
(der. of top)*(bottom) - (top)(der. of bottom)/bottom^2
[1(x-1) - (x+1)(1)]/(x-1)^2
[(2-1) - (2+1)] /(2-1)^2= (1-3)/1 = -2/1
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
`a f ' (x) = [ (x+1)'(x-1) - (x+1)(x-1)'] / (x-1)^2 =
[ (x-1) - (x+1) ] / (x-1)^2 =
-2 / (x-1)^2.
When x = 2 we get f ' (x) = f ' (2) = -2 / (2-1)^2 = -2. **
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Self-critique (if necessary):
n/a
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Self-critique Rating:3
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Question: `q 2 g 7th edition 2.4.34 (was 2.4.30) der of (t+2)/(t^2+5t+6)
What is the derivative of the given function and how did you get it?
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Your solution:
[(der. of top)(bottom)-(top)(der. of bottom)]/(bottom)^2
[(1)((t^2+5t +6) - (t+2)(2t+5)]/(t^2+5t+6)^2
-1(t^2+4t+4)/(t^2+5t+6)^2= [-1(t+2)(t+2)]/[(t+2)(t+3)(t+2)(t+3)]
simplify: -1/[(t+3)^2]
Confidence Rating:3
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Given Solution:
`a we get (f ' g - g ' f) / g^2 = [ 1 ( t^2 + 5 t + 6) - (2t + 5)(t + 2) ] / (t^2 + 5 t + 6 )^2 =
[ t^2 + 5 t + 6 - ( 2t^2 + 9 t + 10) ] / (t^2 + 5 t + 6)^2 =
(-t^2 - 4 t - 4) / (t^2 + 5 t + 6)^2 ) =
- (t+2)^2 / [ (t + 2) ( t + 3) ]^2 =
- 1 / (t + 3)^2.
DER**
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Self-critique (if necessary):
n/a
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Self-critique Rating:3
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Question: `q 4b 7th edition 2.4.48 (was 2.4.44) What are the points of horizontal tangency for(x^4+3)/(x^2+1)?
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Your solution:
[(4x^3)(x^2+1) - (x^4+3)(2x)]/(x^2+1)^2
[4x^5+4x^3 - (2x)(x^4+3)]/(x^2+1)^2
4x^5+4x^3 - 2x^5-6x/(x^2+1)^2 = 2x^5+4x^3-6x/(x^2+1)^2 =
@&
You need parentheses around your numerators in the above.
*@
[2x(2x^2-3)]/(x^2+1)
@&
The step efore this should have read
(2x^5+4x^3-6x)/(x^2+1)^2
It isn't clear how you got from this step to your expression
[2x(2x^2-3)]/(x^2+1)
There is no (x^2+1) factor in the numerator, so the denominator would not have changed.
The numerator is a degree-5 polynomial, and this won't change.
*@
horizontal tangency would be x=0 at x=0 y=0
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
`a the derivative is
( f ' g - g ' f) / g^2 =
(4 x^3 ( x^2 + 1) - (2x) (x^4 + 3) ) / (x^2+1)^2 =
[ 4 x^5 + 4 x^3 - 8 x^5 - 6 x ] / (x^2 + 1)^2 =
-(2 x^5 + 4x^3 - 6x ) / (x^2 + 1)^2 =
2x (x^4 + 2 x^2 - 3) / (x^2+1)^2.
The tangent line is horizontal when the derivative is zero. The derivative is zero when the numerator is zero.
The numerator is 2x ( x^4 + 2 x^2 - 3), which factors to give 2x ( x^2 + 3) ( x^2 - 1).
2x ( x^2 + 3) ( x^2 - 1) = 0 when 2x = 0, x^2 + 3 = 0 and x^2 - 1 = 0.
}2x = 0 when x = 0;
x^2 + 3 cannot equal zero; and
x^2 - 1 = 0 when x = 1 or x = -1.
Thus the function has a horizontal tangent when x = -1, 0 or 1. **
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Self-critique (if necessary):
[ 4 x^5 + 4 x^3 - 8 x^5 - 6 x ] / (x^2 + 1)^2 =
I have worked through this about 5 times and I do not see where you got -8x^5.
I got (2x)(x^4+3) for that side and that gives 2x^5 +6x. Please explain.
@&
That was obviously a typo, since the following step is correct.
Thanks for pointing it out so I could correct it in the original document.
*@
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Self-critique Rating:2
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Question: `qWhat would the graph of the function look like at and near a point where it has a horizontal tangent?
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Your solution:
Points are (-1,2)(0,3)(1,2) looks like a parabola at this point(but it will have more curves elsewhere)
confidence rating #$&*:
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Given Solution:
`a At or near a point of horizontal tangency the graph would become at least for an instant horizontal. This could occur at a peak (like a hilltop, which is level at the very top point) or a valley (level at the very bottom). It could also occur if an increasing function levels off for an instant then keeps on increasing; or if a decreasing function levels off for and instant then keeps decreasing. **
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Self-critique (if necessary):
n/a
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Self-critique Rating:3
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Question: `q 7 7th edition2.4.58 (was 2.4.54) defective parts P = (t+1750)/[50(t+2)] t days after employment
What is the rate of change of P after 1 day, and after 10 days?
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Your solution:
derivative 874/[25(t^2+4t+4)]
rate at t=1 is 3.88
rate at t=10 is .0242
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
`a It doesn't look like you evaluated the rate of change function to get your result.
You have to use the rate of change function to find the rate of change. The rate of change function is the derivative.
The derivative is
( f ' g - g ' f) / g^2 =
( 1 * 50(t+2) - 50(t + 1750) ) / ( 50(t+2) ) ^ 2 =
-50 (1748) / ( 2500 ( t^2)^2 ) =
- 874 / ( 25 ( t + 2) ^ 2 ).
Evaluating when t = 1 and t = 10 we get -3.88 and -.243. **
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Self-critique (if necessary):
Missed the negative. I’m afraid this is just the kind of mistake that’s going to keep me from passing the math Praxis.
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Self-critique Rating:3
@&
Everybody misses negatives from time to time. Just stay aware of this type of error and keep practicing.
*@
(MOSTLY CORRECT) STUDENT SOLUTION TO NON-ASSIGNED PROBLEM WITH INSTRUCTOR COMMENTS IN BOLDFACE:
f(x) = [x^3 + 3x + 2] / (x^2 - 1)
f ' (x) = [(x^2 - 1)(x^3 + 3x + 2)' - (x^3 + 3x + 2) (x^2 - 1) ' ] / (x^2 -1)^2
Since the next step is correct I imagine this is just a typo on your part, but I wanted to be sure to point it out, just in case.
f'(x) = (x^2 - 1)(3x^2 + 3) - (x^3 + 3x + 2) (2x] / (x^2 -1)^2
f'(x) = (3x^4 - 3x^2 + 3x^2 - 3) - ( 2x^4 +6x^2 + 4x)] / (x^2-1)^2
f'(x) = x^4 - 6x^2 -4x - 3 ] / (x^2 - 1)^2
f'(x) = x^4 - 6x^2 - 4x - 3 ] / x^4 - 2x^2 + 1
Except for signs of grouping this is the correct derivative. With signs of grouping it is written
(x^4 - 6 x^2 - 4 x - 3)/(x^2 - 1)^2 or
(x^4 - 6 x^2 - 4 x - 3)/(x^4 - 2 x^2 + 1).
f'(x) = -3 - 4x - 3
f'(x) = -4x - 6
These last two steps are not correct. You can divide out common factors of the numerator and denominator but you can't divide out terms.
For example (4 + 7) / (4 + 3) is not 7/3.
(4 + 7) / (4 + 3) = 11 / 7. The 4's don't divide out.
Neither do the x^4 terms, or the x^2 factors of the -6x^2 and 2x^2 terms of your (correct-up-to-that-point) expression.
You did a very good job taking the derivative. That algebra error isn't one you should be making at this stage, so be sure to take note, but it doesn't detract from the fact that you've done a nice job applying the quotient rule.
"
Self-critique (if necessary):
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Self-critique rating:
(MOSTLY CORRECT) STUDENT SOLUTION TO NON-ASSIGNED PROBLEM WITH INSTRUCTOR COMMENTS IN BOLDFACE:
f(x) = [x^3 + 3x + 2] / (x^2 - 1)
f ' (x) = [(x^2 - 1)(x^3 + 3x + 2)' - (x^3 + 3x + 2) (x^2 - 1) ' ] / (x^2 -1)^2
Since the next step is correct I imagine this is just a typo on your part, but I wanted to be sure to point it out, just in case.
f'(x) = (x^2 - 1)(3x^2 + 3) - (x^3 + 3x + 2) (2x] / (x^2 -1)^2
f'(x) = (3x^4 - 3x^2 + 3x^2 - 3) - ( 2x^4 +6x^2 + 4x)] / (x^2-1)^2
f'(x) = x^4 - 6x^2 -4x - 3 ] / (x^2 - 1)^2
f'(x) = x^4 - 6x^2 - 4x - 3 ] / x^4 - 2x^2 + 1
Except for signs of grouping this is the correct derivative. With signs of grouping it is written
(x^4 - 6 x^2 - 4 x - 3)/(x^2 - 1)^2 or
(x^4 - 6 x^2 - 4 x - 3)/(x^4 - 2 x^2 + 1).
f'(x) = -3 - 4x - 3
f'(x) = -4x - 6
These last two steps are not correct. You can divide out common factors of the numerator and denominator but you can't divide out terms.
For example (4 + 7) / (4 + 3) is not 7/3.
(4 + 7) / (4 + 3) = 11 / 7. The 4's don't divide out.
Neither do the x^4 terms, or the x^2 factors of the -6x^2 and 2x^2 terms of your (correct-up-to-that-point) expression.
You did a very good job taking the derivative. That algebra error isn't one you should be making at this stage, so be sure to take note, but it doesn't detract from the fact that you've done a nice job applying the quotient rule.
"
Self-critique (if necessary):
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Self-critique rating:
#*&!
&#This looks good. See my notes. Let me know if you have any questions. &#