[maths] Re: more of calculus
- From: "Nelson Blachman" <blachman@xxxxxxx>
- To: <maths@xxxxxxxxxxxxx>
- Date: Fri, 01 Dec 2006 22:38:58 -0800
Ned,
Your solution to the earlier Problem 3:
Find the absolute maximum and minimum of f on the given interval:
3. q: f(x) = x^3 - 3x + 1, [0,3].
seems pretty good. I guess you differentiated f(x), getting
f'(x) = 3x^2 - 3,
which vanishes when x = +-1.
The second derivative is
f"(x) = 6x,
which is positive and negative at these 2 values of x, respectively, showing
thet f(x) has a local maximum at x=-1
and a local minimum at x=1, viz., -1, which happens to be the "absolute
minimum" on the interval [0,3].
I hope you have a mental image of this typical cubic function. On this
interval the "absolute maximum" occurs where x=3, viz., 19.
This approach should be helpful to you in dealing with the succeeding problems.
I regret that I don't recall just what the theorems mentioned there state.
HTH
Nelson
----- Original Message -----
From: Ned Granic
To: maths@xxxxxxxxxxxxx
Sent: Monday, November 27, 2006 5:14 PM
Subject: [maths] more of calculus
Hi all,
Hope you're doing good, good people.
I have a list of questions that are marked as even problems in my homework
which I doubt the results of, so I'll post the problems and the so-called
solutions I kinda found, and if somebody cares to verify if they are
accidentally the right ones.
Find the critical numbers of the function:
1. q: f(x) = x^3 + x^2 -x.
a: 1/3, -1.
2. q: f(x) = x^3 + x^2 + x.
a: Not found in R;
the imaginary solution I found is (-1+i*sqrt(2))/3.
(we don't do imaginary solutions, that was my only escape).
Find the absolute maximum and minimum of f on the given interval:
3. q: f(x) = x^3 - 3x + 1, [0,3].
a: abs max is f(0) = 1, f(3) = 1;
abs min is f(1) = -1.
4. q: f(x) = x^3 - 6x^2 + 9x + 2, [-1,4].
a: abs min is f(-1) = -14;
abs max is f(1) = 6, f(4) = 6.
Verify that the function satisfies the 3 hypotheses of Rolle's Theorem on the
given interval. Then find all numbers c that satisfy the conclusion of that
theorem.
5. q: f(x) = x^3 - 3x^2 + 2x + 5, [0,2].
a: Well, the function is a polynomial, therefore continuous and
differentiable,
and c values are 1 +or- 1/sqrt(3).
Verify that the function satisfies the hypothesis of the Mean Value Theorem
on the given interval. Then find all numbers c that satisfy the conclusion of
the theorem.
6. q: f(x) = x^3 + x - 1, [0,2].
a: c = 7/6; -7/7 is not in the domain of f.
7. f(x) = x/(x+2), [1,4].
a: the c number I've got is (-4+-2*sqrt(3))/2
= (-2+sqrt(3)). <cough, cough>
Enough for now, more to come. Sttay tune!
And, of course, many thanks in advance!
Ned
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