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

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In summary, the conversation is about finding the values of a, b, and c in the function f(x)= ax3 + bx2 + cx, given that it has a point of inflection at the origin and a local maximum at the point (2,4). The process involves using the values of the point of inflection and the local maximum to create three equations and solve for the unknown variables. There was some discussion about the correct values for b and c, but ultimately it was determined that b=0 and c=0, and a=-1/4. The final function is f(x)=-x^3/4+3x.
  • #1
bengalibabu
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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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  • #2
I would start by differentiating what you have.
 
  • #3
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.
 
  • #4
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?
 
  • #5
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
 
  • #6
what am i doing wrong here, is b=0?
 
  • #7
since the inflection point is x=0 b should be 0 if your work is correct. ur second derivative is right.
 
  • #8
im just really being hesitant on the fact that since b=0 then c must also equal 0
 
  • #9
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

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

therefore

[tex] f(x)=-\frac{x^{3}}{4}+3x [/tex]

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
 

What is a point of inflection?

A point of inflection is a point on a graph where the concavity changes from upward to downward or vice versa. It is the point where the second derivative of the function is equal to zero.

How do you find the coordinates of a point of inflection?

To find the coordinates of a point of inflection, you need to first find the second derivative of the function and set it equal to zero. Then, solve for the x-value of the point of inflection. Plug in this x-value into the original function to find the corresponding y-value.

What is a local maximum?

A local maximum is the highest point on a graph within a certain interval. It is also known as a relative maximum. It is the point where the function changes from increasing to decreasing.

How do you find the coordinates of a local maximum?

To find the coordinates of a local maximum, you need to first take the first derivative of the function and set it equal to zero. Solve for the x-value of the local maximum. Plug in this x-value into the original function to find the corresponding y-value.

Can a point of inflection and a local maximum be the same point?

Yes, it is possible for a point of inflection and a local maximum to be the same point. This can happen if the function has a horizontal point of inflection, where the second derivative is equal to zero, and a local maximum at the same x-value. However, this is not always the case and it is not a common occurrence.

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