Okay, so I have a calc problem:
Directions say this problem has one local max and one local min point.
f(x)= (-1/3)x³+2x²-12
Math I've done:
d/dx = 3(-1/3)x²+4x
d/dx = -x²+4x
0 = -x²+4x
0/-x = -x²+4x / (-x)
0 = x - 4
so x = 4
then
d²/dx² = -2x+4 | x=4
= -2(4)+4 = -4 < 0
so concave down, with local max @ x=4, and no local min and no inflection point.
BUT if I don't simplify 0 = -x²+4x ... x = 0 or 4. and then:
d²/dx² = -2x+4 | x=0,4
= -2(0)+4 = 4 > 0
= -2(4)+4 = -4 < 0
and then there is a local min @ x=0 and a local max @ x=4
then:
d²/dx² = -2x+4
0 = -2x+4
-4 = -2x.. -4/(-2) = -2x/(-2)
2 = x ... so inflection point at x = 2.
The book has the bottom answer and I'm sure its correct..
All I want to know is: What am I missing? Why do I get the wrong answer? I've divided both sides by (-x), shouldn't it still work?
If the problem were: 0 = 3x²-27 .. I'd divide both sides by 3, and still get the right answer.
Why not when i divide by x?
2006-09-13 08:29:27 · 7 answers · asked by Sean06 2 in Science & Mathematics ➔ Mathematics
when you do 0/x you're dividing 0 by x. That is valid. IF it were x/0 that would by invalid. Agree? or Disagree?
2006-09-13 08:42:33 · update #1
Sorry about the last "additional details" I read the response incorrectly :) lol just disregard it.
Okay, Is there a mathematical explanation as to why it is eliminating the possibility that x=0 or am I just going to have to see the graph to realize it?
2006-09-13 08:45:47 · update #2
Ah okay now I see what you're saying.
Thanks! =D
2006-09-13 08:50:21 · update #3
OKAY. I really see where I went wrong.
The reason I wanted to divide both sides by -x was to make it easier to factor -x²+4x.
DUH
I took out the x... I should have made it -x(x+4).
and then that is where the x=0, x=4 comes in.
Wow..
2006-09-13 09:00:48 · update #4
Everytime you divide by a variable you lowe the degree of the equation. Every degree represents an answer. A three degree polynomial will have three answers.
2006-09-13 09:33:01 · answer #1 · answered by Anonymous · 1⤊ 0⤋
How can you divide by x when x=0? Your assumption you can is erroneous. It's not a matter of simplifying, it's a matter of performing valid operations.
Yes.. but in an expression such as
x(x-4)=0 you may only divide by x when x<>0 and you may only divide by (x-4) when x<>4. x is not a value it's a variable, the whole domain must be considered!
By your logic 0*-4=0 implies -4=0
2006-09-13 15:38:09 · answer #2 · answered by Andy S 6 · 0⤊ 0⤋
you are not supposed to divide sides by the variable....since you may eliminate possibilities...
consider: x^2 - 6x = 0 then as we factor LHS as x(x-6)=0 we get two possibilities: x=0 or x=6
when you divide sides by x, you are in fact eliminating the possibility that x=0
if x^3-4x^2=0 then x^2(x-4)=0 and we have THREE values for x: 0, 0, 4 (0 doubles)...when you make a graph of the function you will understand the implication of it...
2006-09-13 15:43:15 · answer #3 · answered by m s 3 · 0⤊ 0⤋
I fully agree with the first answerer.
2006-09-13 15:35:01 · answer #4 · answered by KCD 4 · 0⤊ 2⤋
Are you serious? What are you, a rocket scientist?
Showing off now, aren't you.
2006-09-13 15:37:01 · answer #5 · answered by Anonymous · 0⤊ 1⤋
What the hell? Where do you go to school?
2006-09-13 15:48:48 · answer #6 · answered by Caitlyn S 2 · 0⤊ 0⤋
maybe because you're dividing numbers by letters . . .
2006-09-13 15:33:21 · answer #7 · answered by Anonymous · 0⤊ 2⤋