Wordodd --
1. 13-3x is indeed correct
2. x less than or equal to 4
3. correct: all real numbers
4. since A = Pi * r^2 -- you have to replace r with d/2; thus A = (Pi * d^2)/4
5. this function is many things:
a. DECREASING (negative slope)
b. has a slope of -3
c. a linear function
6. rewrite the function: y^2 = -x^2 - 2x + 15
next: do a quick graph (you would expect a downward-opening parabola
just from its equation)
x = 4; y^2 = -16 - 8 + 15; undefined
x = 3; y^2 = -9 - 6 + 15; y = 0 (and so is y>3 is undefined)
x = 2; y^2 = 7; y = 2.65
x = 1; y^2 = 12; y = 2.46
x = 0; y^2 = 15; y = 3.87
x = -1; y^2 = 16; y = 4
x = -2; y^2 = 15; y = 3.87
x = -4; y ^2 = -16 + 8 + 15; y = 2
x = -5; y^2 = -25 + 10 + 15; y = 0
x = -6; y^2 = -36 + 12 +15; undefined (and so is y<-5)
This function is symmetric about a negative number. (We'll let you
wait until you get to calculus to figure out at what point the slope
of this parabola is zero. . .)
7. This function is increasing on the interval of x < 0. (Again, a
quick graph should convince you -- but remember that your binomials
with x^2 are always parabolas.) F(x) is POSITIVE on the interval that
you've defined.
8. g(x) = f (x+3) -2 -- you're going in the X direction +3 units; -2
units in the Y direction
9. You had to send me back to the text books for this one -- and it's
probably important to know the definition of BOTH even and odd
functions.
Illinois Math & Science Academy
"Some Explorations with Even and Odd Functions" (Hamberg, undated)
http://staff.imsa.edu/math/journal/volume4/articles/ExploreEvenOdd.pdf
With an ODD function f(x) = - f(-x) for all values in the domain. --
OR perhaps easier:
f(x) + f(-x) = 0 for all values in the domain.
That's only true of f(x) = -x^3
10. x^2 and x^2 + 1 are the same parabola -- with one shifted up 1
unit in the Y direction
** 11. You're going to have to explain this one a little more to me. .
. what's the relationship of h to the function?
12. Moving right on the X axis is positive; left is negative -- so
this is (x-2). If the factor of 3 is used, it's A
** 13. Is this one f(x) = -x^3 + 4x^2 +16x + 8 or the following (and
it obviously it makes a big difference):
f(x) = x^3 + 4x^2 +16x + 8
14. h(x) = FRAC (2x-1/x-2); so h(h(x) = FRAC (2x-1/x-2)^2
13. We'll work each of these out:
A. (fog)(x) = 4x + 2 + 1 = 4x + 3
B. (fog)(x) = 2x^2 + 1
C. (fog)(x) = (2x+1)^2
D. (fog)(x) = 4x^2 + 1
so it's C
16. B, 5 units left
17: Substitute CUBEDROOT(x) for x^3 -- f(x) = (x^1/3)^3 + 27 = x + 27
If you can clarify the two with ** next to the numbers, we can
probably nail them all down!
Google search strategy:
"odd function" + math + definition
Best regards,
Omnivorous-GA |
Request for Answer Clarification by
wordodd-ga
on
23 Mar 2004 05:52 PST
Ok, i omitted a bit on question 11 let me retype it.
11. 11. If h does not equal 0 and f(x) = x^2 + 3 then the diff. quotient
simplifies to ___________
You needed to plug the f(x) into the diff. quotient - FRAC f(x + h) - f(x)/h
so you plug in x^2 + 3 into that formula
13. Is this one f(x) = -x^3 + 4x^2 +16x + 8 or the following (and
it obviously it makes a big difference):
f(x) = x^3 + 4x^2 +16x + 8
You are right no - before the first term so the question is supposed to be
13. If f(x) = x^3 + 4x^2 + 16x + 8 and g(x) = 4 then (gof)(8) = ____________
a. 3616
b. 4
c. -4
d. none
One last question I didn't understand your answer for it...
10. The function which is one to one is
a. f(x) = x^3 + x + 1
b. f(x) = x^2
c. f(x) = x^2 + 1
d. f(x) = 5
Which ONE of these is a one to one function? Maybe I didn't understand
your answer but it didn't seem to pick just one of them.
Thank you for your prompt inital response. I will take care of the tip
when we finish this up.
|
Request for Answer Clarification by
wordodd-ga
on
23 Mar 2004 05:59 PST
Also, number 7
7. The function f(x) = -x^2 + 1 is increasing on the interval _________
I say ( -1 , 1 )
The choices are
a. ( - infinity, 0 )
b. ( 0 , infinity )
c. ( -1 , 1 )
d. ( 1 , infinity )
|
Clarification of Answer by
omnivorous-ga
on
23 Mar 2004 07:45 PST
Wordodd --
Here goes!
7. is increasing on the area x<0, so it's all of the negative numbers -- choice A
10. Here's where I made a mistake: B and C are identical functions as
their variables are identical (and C only has a constant of 1 added).
However this is a precise definition of a one-to-one function -- and
it's got a GREAT Java applet for graphing:
AnalyzeMath
"One-to-One Function Tutorial"
http://www.analyzemath.com/OneToOneFunct/OneToOneFunct.html
Basically a one-to-one function has no more than 1 input value
producing the same output value. Or, phrased another way, no two
numbers X can produce the same f(X). So that leaves out the
binomials, because they're symmetric parabolas: -1 and 1 produce the
same result for the functions in both B and C.
If f(x) = 5 for any number x -- that one's excluded too -- f(anything) is a 5!
Only the function in A is a one-to-one or "injective" function.
Use the graphing capability on the page above to see why it produces a
unique result for each number x -- and the other 3 don't.
11. You've got me on this one -- and I think that it's a confusion
over terms. Is "diff. quotient" indeed the "differential quotient"?
And FRAC a "fractional part" function? It appears that you're trying
to calculate a derivative for the function f(x) = x^2 + 3
13. g(x) = 4 for any number x . You could put an avocado in there
and the answer would be 4. Putting another function into a constant
produces 4.
Answer: B
Best regards,
Omnivorous-GA
|
Request for Answer Clarification by
wordodd-ga
on
23 Mar 2004 08:12 PST
Yes, you are correct differential quotient.
FRAC was just stating that what followed was a fraction, I did that to
try to make it easy to understand. Either way I am fairly sure it
reduces to FRAC h/h or h but if I am wrong please tell me.
I turn the test in Wednesday and probably get a grade in class, so
look for your tip Wednesday afternoon if he takes them home to grade
then I will tip you the following Monday after class. Thank you for
all the help, I will rate you the highest I can.
|
Clarification of Answer by
omnivorous-ga
on
23 Mar 2004 09:26 PST
Wordodd --
This number 11 still has me confused. I don't want to introduce the
definition of a derivative from calculus because it may be taking this
in the wrong direction -- but that's where a differential quotient is
typically used.
I'm not familiar with a differential quotient defined as: FRAC f(x+h) - f(x)/h
Without a better definition of the FRAC function, it could return any number.
Best regards,
Omnivorous-GA
|
Request for Answer Clarification by
wordodd-ga
on
23 Mar 2004 10:15 PST
11. If h does not equal 0 and f(x) = x^2 + 3 then the diff. quotient
simplifies to ___________
f(x+h) - f(x)/h
how about this (f(x+h) - f(x)) / h
(x^2 + 3 + h - (x^2 + 3)) / h
(x^2 + 3 + h - x^2 - 3) / h
(h) / h = h or 1 but 1 isn't a choice
|
Clarification of Answer by
omnivorous-ga
on
23 Mar 2004 11:20 PST
Wordodd --
Is 2h a choice? That's what the derivative would be. . .
Best regards,
Omnivorous-GA
|
Request for Answer Clarification by
wordodd-ga
on
23 Mar 2004 11:36 PST
no the choices are
a. 2x + h
b. h
c. h + 3
d. 3x
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Clarification of Answer by
omnivorous-ga
on
23 Mar 2004 17:42 PST
Wordodd --
Let's try #11 this way (and this may be a little different way of
looking at the problem than you've seen with differential quotients).
With the function f(x) = x^2 + 3, we're going to go from any point x
to a point where x = 1+h
We'll have the following numbers:
x2 = x+h
x1 = x
y2 = (x+h)^2 + 3
y1 = x^2 + 3
the differential quotient (or slope of a straight line between the two points is):
(y2 - y1)/(x2 -x1), where x2-x1 = h
and, y2 -y1 = (x+h)^2 + 3 - (x^2 + 3) = x^2 + 2xh + h^2 + 3 - x^2 - 3 =
2xh + h^2
Thus the answer should be h(2x+h)/h = 2x +h
ANSWER: A
In fact, this is a case of using differential quotients to come up
with the first derivative or slope of a curve.
Best regards,
Omnivorous-GA
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