[maths] Re: antiderivatives; calculus
- From: "Nelson Blachman" <blachman@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Fri, 01 Dec 2006 17:57:24 -0800
Ned,
You began with the right f(x)= x + 1/x, but you should not have set it equal
to zero, and having set it equal to zero, you didn't solve the resulting
equation correctly; the correct solution to that equation is x = i or -i, where
i is either of the two square roots of -1.
If you'd differentiated f(x), you'd have gotten 1 - 1/x^2 = 0, whose
solutions are, of course x = 1 or -1. Then you'd have to look at the value of
f(1) and f(-1) to see which, if either, of these is a minimum of f(x).
Since the statement of the problem excludes x=-1, you need only to look at
x=1, where the second derivative of f(x) is 2/x^3. Since this is positive for
x=1, we can conclude that the point (1,2) is the minimum of f(x) insofar as
positive values of x are concerned; if you allow x<0, there's no global
minimum, only this local minimum at x=1 as well as a local maximum at x=-1,
since f(x) can be arbitrarily small when x gets arbitrarily negative.
I've been planning to look again at your earlier e-mail to consider problems
3 & 4 when I've found time to clean up the seventy unread e-mails now in my
Inbox.
--Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Friday, December 01, 2006 5:15 AM
Subject: [maths] Re: antiderivatives; calculus
Thanks Nelson again!
To be honest, it's from you that I hear of the term "integral" for the first
time, and they are in my next chapter.
I have one more problem that I'm uneasy about:
Find a positive number whose reciprocal added to it is as small as possible.
I've got 1.
x+(1/x) >= 0
is it correct?
Also, if you don't mind verifying the results I posted in my first or second
e-mail this week.
Many thanks for everything!
Ned
----- Original Message -----
From: Nelson Blachman
To: maths@xxxxxxxxxxxxx
Sent: Friday, December 01, 2006 12:24 AM
Subject: [maths] Re: antiderivatives; calculus
Ned,
The indefinite integral of f(x)=x^(3/4) + x^(4/3) is
F(x) = (4/7)x^(7/4) + (3/7)x^(7/3) + C,
just as you said. Congratulations! I hope you can check such results
yourself by differentiating F(x) and getting f(x).
I wonder why your teacher uses the term "antiderivative" when the
standard term is "integral," which is half as long.
--Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Thursday, November 30, 2006 2:28 PM
Subject: [maths] antiderivatives; calculus
If f(x) = fourth_root(x^3) + cube_root(x^4), then
its most general antiderivative function, I thinks, is:
F(x) = 4/7x^7/4 + 3/7x^7/3 + C.
How far off am I, that is the question!
Cheers!
Ned
Other related posts:
