# quick Calc Equation question

• Dec 29th 2008, 05:03 PM
OnMyWayToBeAMathProffesor
quick Calc Equation question
Hello,
I am stuck with this one question, I do not even know how to begin. Any help would be greatly appreciated.

Determine \$\displaystyle a\$, \$\displaystyle b\$, \$\displaystyle c\$ and \$\displaystyle d\$ so that the graph of \$\displaystyle ax^3+bx^2+cx+d\$ has a point of inflection at the origin and a relative maximum at the point \$\displaystyle (2,4)\$.

I know the point of inflection is when the second derivative equals 0 and the max is when the first derivative equals 0, going from negative y values to positive y values. but how do i use that?

• Dec 29th 2008, 05:15 PM
TheMasterMind
Quote:

Originally Posted by OnMyWayToBeAMathProffesor
Hello,
I am stuck with this one question, I do not even know how to begin. Any help would be greatly appreciated.

Determine \$\displaystyle a\$, \$\displaystyle b\$, \$\displaystyle c\$ and \$\displaystyle d\$ so that the graph of \$\displaystyle ax^3+bx^2+cx+d\$ has a point of inflection at the origin and a relative maximum at the point \$\displaystyle (2,4)\$.

I know the point of inflection is when the second derivative equals 0 and the max is when the first derivative equals 0, going from negative y values to positive y values. but how do i use that?

Let f(x) = ax3 + bx2 + cx + d
f'(x) = 3ax2 + 2bx + c
f''(x) = 6ax + 2b

f(2) = 8a + 4b + 2c + d = 8a + 2c = 4
f'(2) = 12a + 4b + c = 12a + c = 0

Solve

8a + 2c - 2(12a + c) = 4 - 0
-16a = 4
a = -1/4

12(-1/4) + c = 0
c = 3

Therefore, f(x) = -1/4 x3 + 3x
• Dec 29th 2008, 05:16 PM
nzmathman
This question was answered yesterday here: http://www.mathhelpforum.com/math-he...213-b-c-d.html