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FAO: Math People (Need Help w/ Calculus)
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| TruffleShuffle |
I used to know how to do this problem a long time ago, but I can't find my notes on it:
If the graph of y = x^3 + ax^2 + bx - 4 has a point of inflection at (1,-6), what is the value of b?
a) -3
b) 0
c) 1
d) 3
e) It cannot be determined from the information given.
Anyone here who can help me? Thanks in advance.
Edit: Here's another one I need help with, that I've never seen before:
The general solution of the differential equation y'= y + x^2 is y =
a) Ce^x
b) Ce^x + x^2
c) -x^2 - 2x - 2 + C
d) e^x - x^2 - 2x - 2 + C
e) Ce^x - x^2 - 2x - 2
Help with either of these or both would be greatly appreciated. Thanks! |
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| Dr. Cfire |
y = x^3 + ax^2 +bx -4
y' = 3x^2 +2ax + b
y'' = 6x + 2a
since point of inflection at (1,-6)
0 = 6(1) +2a
therefore a = -3
now we know that the point (1,-6) is on the line
y = x^3 + 2(-3)x + b
-6 = 1^3 + 2(-3)(1) + b
therefore b = -1 himm I think I ed it up
:mad: |
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| dr me |
i'm bored so i'll help
i'm sure there's a better way to do this but its been going on 8yrs since i did maths
y = x^3 + ax^2 + bx - 4 (1)
y' = 3x^2 + 2ax + b (2)
since y'=0 at x=1 then by substituting x=1 and y'=0 into (2) we get
b = -2a - 3 (3)
now substitute (3) into (1)
y = x^3 + ax^2 - (2a+3)x - 4 (4)
then solve (4) using (1,-6) and you get a=0
then solve (1) using (1,-6) with a=0 and you get b=-3
actually that was mainly algebra with kiddy calculus
Edit: with the second one just differentiate each of the solutions until you get your answer |
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| TruffleShuffle |
I think Cfire had it right; I went back and plugged the 'a' and 'b' values into the original equation and went to find the point of inflection myself.
By the way, I find it kind of interesting how both your names begin with "Dr" |
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| Dr. Cfire |
| quote: | Originally posted by dr me
i'm bored so i'll help
i'm sure there's a better way to do this but its been going on 8yrs since i did maths
y = x^3 + ax^2 + bx - 4 (1)
y' = 3x^2 + 2ax + b (2)
since y'=0 at x=1 then by substituting x=1 and y'=0 into (2) we get
b = -2a - 3 (3)
now substitute (3) into (1)
y = x^3 + ax^2 - (2a+3)x - 4 (4)
then solve (4) using (1,-6) and you get a=0
then solve (1) using (1,-6) with a=0 and you get b=-3
actually that was mainly algebra with kiddy calculus
Edit: with the second one just differentiate each of the solutions until you get your answer |
Umm the roots of the first derivative is the maximum and minimum not the points of inflection. The roots of the second derivative is the points of inflection. |
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| PersianMafia |
| quote: | Originally posted by TruffleShuffle
I used to know how to do this problem a long time ago, but I can't find my notes on it:
If the graph of y = x^3 + ax^2 + bx - 4 has a point of inflection at (1,-6), what is the value of b?
a) -3
b) 0
c) 1
d) 3
e) It cannot be determined from the information given.
Anyone here who can help me? Thanks in advance.
Edit: Here's another one I need help with, that I've never seen before:
The general solution of the differential equation y'= y + x^2 is y =
a) Ce^x
b) Ce^x + x^2
c) -x^2 - 2x - 2 + C
d) e^x - x^2 - 2x - 2 + C
e) Ce^x - x^2 - 2x - 2
Help with either of these or both would be greatly appreciated. Thanks! |
AP exam is coming up soon isn't it. Yeah I got a mock this Sunday...
:nervous: |
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| THE_Chris |
Be VERY thankful that I did this :) Always useful to have a recap on my old stuff when Im going into my final exams....
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