Calc Help: Find a, b & c Using Point of Inflection & Local Max

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The discussion revolves around finding the coefficients a, b, and c in the cubic function f(x) = ax^3 + bx^2 + cx, given a point of inflection at the origin and a local maximum at (2,4). Participants emphasize the need to differentiate the function to derive equations based on the conditions provided. The point of inflection indicates that the second derivative is zero at (0,0), while the local maximum provides two equations: f(2) = 4 and f'(2) = 0. Through calculations, it is concluded that a = -1/4, b = 0, and c = 3, leading to the function f(x) = -1/4x^3 + 3x. The discussion highlights the importance of correctly applying calculus principles to solve for unknown coefficients.
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Calculus help please!

The function f(x)= ax3 + bx2 + cx has a point of inflection at the origin and a local maximum at the point (2,4). Find the values of a, b and c.

I understand that the point of inflection is (0,0) and the local maximum at (2,4) but how can u find a, b & c using these values?
 
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I would start by differentiating what you have.
 
You need three equations for your three unkowns. The point of inflection indicates that the second derivitive is zero at (0,0). The local maximum at (2,4) gives you two pieces of information which can give you equations. First of all, f(2) must be 4 in order for the graph to pass through this point. Secondly the derivitive is zero at two. This gives you three equations in three unkowns, so you can solve them.
 
Hint:

What do f'(x) and f''(x) tell you about points of inflection, local maxima etc.?

Can you find f'(x) and f''(x) here, at the points x=0 and x=2?
 
so is this how it is done:

f(x)= ax3 + bx2 + cx

4=a(2)^3 + b2^2 + 2c
4=8a + 4b +2c
4=2(4a + 2b + c)
2=(4a + 2b + c)

0=3ax^2 + 2b^2 +c0
0=3a(4) + (4)b + c
c= -12b - 4b

6ax + 2b = 0
b=-3ax
b=0

f(x)= ax3 + bx2 + cx
2=(4a + 2b + c)
2=4a
a=1/2
 
what am i doing wrong here, is b=0?
 
since the inflection point is x=0 b should be 0 if your work is correct. ur second derivative is right.
 
im just really being hesitant on the fact that since b=0 then c must also equal 0
 
but does a=1/2
 
  • #10
well if b and c equal 0 then a = 1/2
 
  • #11
ok b does equal 0 but what about c
 
  • #12
k nelson...if b=o then c=0 as well right according to the calculations...or is that wrong too?
 
  • #13
I get

\left\{\begin{array}{c} a=-\frac{1}{4} \\ b=0 \\ c=3,

therefore

f(x)=-\frac{x^{3}}{4}+3x

Daniel.
 
  • #14
yes, the calculation you may have made a minor error


0=3a(4) + (4)b + c
c= -12b - 4b

yes c would equal zero according to this statement

but i think you meant
c=-12a - 4b
 
  • #15
thx for ur help but dexter was write...thx guys
 
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