| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
I am trying to help my daughter with a calculus problem, and I have no
clue...
Given the following problems, and there are two... I need to be able to
find:
1) the X and Y intercepts.
2) the values when the 1st derivative is equal to zero
3) the values when the 2nd derivative is equal to zero
(we expect the answers to be 0, 1 and -1, but we don't know how to show
the work to get there.... can you help.
And... please show us the work... and reasoning behind it... :-)
Need to know, how we made them = to zero, and why....
1st problem
2/3 2
3x - x = y
-1/3
2x - 2x == is what we believe is the 1st derivative
-4/3
-2/3x - 2 == is what we belive is the 2nd derivative
Can you verify the work of the derivative...
2nd problem
2
(x - 7x + 13)/ (x-4)
2
(x-5) (x-3)/(x-4) == what we belive is the 1st derivative
2/(x-4) === the daughter says after "lots" of simplification: 2nd deriv
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 2016.1 | CSC32::D_DERAMO | Dan D'Eramo, Customer Support Center | Fri Dec 01 1995 03:17 | 16 | |
> 2
> (x - 7x + 13)/ (x-4)
Since (x-4)(x-3) = x^2 - 7x + 12, you can write the above
as
x - 3 + (1/(x-4))
or
-1
x - 3 + (x - 4)
That should make computing the derivative easier.
Dan
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| 2016.2 | AUSSIE::GARSON | achtentachtig kacheltjes | Fri Dec 01 1995 04:15 | 15 | |
re .0
To find the x-axis intercept, set y = 0 and solve for x.
Remember that there can be more than one x-axis intercept.
To find the y-axis intercept, set x = 0 and solve for (i.e. evaluate) y.
You appear to have correct first and second derivatives for the first
function so go ahead and solve for their being 0.
Based on the hint in .1 you appear to have one or more errors in the
work for the second function.
P.S. We tend not just to do homework problems for people.
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| 2016.3 | Help and Hint is terrific, but need more help | WROSS1::LIBATIQUE_LO | Fri Dec 01 1995 23:09 | 23 | |
Thanks for the help...and it has helped, especially .1
However, did I tell you that I was 2000 miles from my daughter, and it
was important for me (not perhaps you) to help her with her homework.
The self-help is terrific, and I understand the nature of giving
"hints", but I would like to help her answer her problems... :-)
I'm sure you understand what it must be like to be the occasional
remote mom/dad. If someone wants to help show the work through, that
would be great.
I had forgotten the idea of taking x^2 -7x +13 ---> x^2-7x+12+1
-2
SO that leaves me with a 1st der: -1(x-4) + 1 = dx/dy...right?
-3
Accordingly, the 2nd der would be 2(x-4) =ddx/dy...right?
See...I am trainable... now to look at both problems again...because
not only did I need the intercepts, also the max/min/asymetrical line
?? points, etc...some of which I just don't remember when 2000 miles
from child and her book...
jff
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| 2016.4 | AUSSIE::GARSON | achtentachtig kacheltjes | Sat Dec 02 1995 03:21 | 4 | |
re .3
Yes, I agree with your result for the first and second derivative of
the second function.
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| 2016.5 | MAL009::ROOS | Wed Dec 06 1995 18:51 | 27 | ||
First example:
1st derivative is correct:
2nd derivative is correct:
Now, when does each equal zero?
1st derivative equals zero when x = 1 or -1
2nd derivative never equals zero - it is always a negative value
Second example:
ist derivative is correct
2nd derivative is equal to 2/(x-4) to the 3rd power - third power in
the denominator
Now, when does each equal zero?
1st derivative equals zero when x = 3 or x = 5
2nd derivative never equals zero - when x > 4 it is positive; when x <
4 it is negative
Got any more?
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