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!

Last edited by Psionic on Apr-23-2005 at 02:55

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 fucked it up

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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huh?

Last edited by dr me on Apr-23-2005 at 04:55

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"

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.

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...

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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