# Calculus help !

bengalibabu

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?

noslen
I would start by differentiating what you have.

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.

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?

bengalibabu
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

bengalibabu
what am i doing wrong here, is b=0?

punjabi_monster
since the inflection point is x=0 b should be 0 if your work is correct. ur second derivative is right.

bengalibabu
im just really being hesitant on the fact that since b=0 then c must also equal 0

noslen
but does a=1/2

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

noslen
ok b does equal 0 but what about c

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

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.

noslen
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

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