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You are here: Experts > Teens > Homework/Study Tips > Calculus > Basic curve sketching with derivatives.
Expert: Paul Klarreich
Date: 5/12/2008
Subject: Basic curve sketching with derivatives.
Question
QUESTION: 2) Consider F(x)=x^3+ax^2+bx+c, where a,b, and c are constants with a>0(or equal to 0) and b>0 (or equal to 0).
a) Use second derivatives to explain its concavity and inflection point.
b) If (5,7) is its inflection point, then find (a) and (c) and its critical point if there is any.
ANSWER: Questioner: David
Category: Calculus
Private: No
Subject: relative max & min problem
Question: 2) Consider F(x)=x^3+ax^2+bx+c, where a,b, and c are constants with a>=0 and b>=0
a) Use second derivatives to explain its concavity and inflection point.
b) If (5,7) is its inflection point, then find a and c and its critical point if there is any.
..........................................
Hi, David,
F' = 3x^2 + 2ax + b
F'' = 6x + 2a.
There would be an inflection point when F'' = 0,
6x + 2a = 0
x = -a/3
If x < -a/3, then F'' < 0 ==> concave down,
If x > -a/3, then F'' < 0 ==> concave up.
If (5,7) is the inflection point, then
5 = -a/3, so a = - 15
You have now:
F(x) = x^3 - 15x^2 + bx + c
F(5) = 125 - 15(25) + 5b + c
F(5) = 125 - 375 + 5b + c
F(5) = -250 + 5b + c = 7
So 5b + c = - 243
Any ideas about b? I think you need some more info.
---------- FOLLOW-UP ----------
QUESTION: What about its critical point?
Answer
Questioner: David
Private: no
Subject:
Question:
QUESTION: 2) Consider F(x)=x^3+ax^2+bx+c, where a,b, and c are constants with a>0(or equal to 0) and b>0 (or equal to 0).
a) Use second derivatives to explain its concavity and inflection point.
b) If (5,7) is its inflection point, then find (a) and (c) and its critical point if there is any.
ANSWER: Questioner: David
Category: Calculus
Private: No
Subject: relative max & min problem
Question: 2) Consider F(x)=x^3+ax^2+bx+c, where a,b, and c are constants with a>=0 and b>=0
a) Use second derivatives to explain its concavity and inflection point.
b) If (5,7) is its inflection point, then find a and c and its critical point if there is any.
..........................................
Hi, David,
F' = 3x^2 + 2ax + b
F'' = 6x + 2a.
There would be an inflection point when F'' = 0,
6x + 2a = 0
x = -a/3
If x < -a/3, then F'' < 0 ==> concave down,
If x > -a/3, then F'' < 0 ==> concave up.
If (5,7) is the inflection point, then
5 = -a/3, so a = - 15
You have now:
F(x) = x^3 - 15x^2 + bx + c
F(5) = 125 - 15(25) + 5b + c
F(5) = 125 - 375 + 5b + c
F(5) = -250 + 5b + c = 7
So 5b + c = - 243
Any ideas about b? I think you need some more info.
---------- FOLLOW-UP ----------
QUESTION: What about its critical point?
-------------------------------------------
To find the C.P. set F' = 0, and write:
3x^2 + 2ax + b = 0
Any ideas about b?
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