# Calculus help !

Calculus help plz!

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.

LeonhardEuler
Gold Member
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.

James R
Homework Helper
Gold Member
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

well if b and c equal 0 then a = 1/2

ok b does equal 0 but what about c

k nelson....if b=o then c=0 as well right according to the calculations...or is that wrong too?

dextercioby
Homework Helper
I get

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

therefore

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

Daniel.

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

thx for ur help but dexter was write...thx guys