If f(x)=ax^3+bx^2+cx+d, a ≠ 0, determine...?

Question

If f(x)=ax^3+bx^2+cx+d, a ≠ 0, determine a,b,c,d so that f has a local minimum at (0,3) and a local maximum at (4,12).

I understand that the critical numbers are 0 and 4, but I'm not quite sure how to start this problem. Any help/work is much appreciated. Thanks.

Answer 1

you will have a system of equations...

f(0) = 3 ... then d = 3
f'(0) = 0
f(4) = 12
f'(4) = 0

you will have by this time
3 linear equations in 3 variables: a, b & c.

you can thus solve for them...


§

Answer 2

Whenever you are asked to find minimum or maximum, first thing to do is find the derivative

f '(x) = 3ax^2 +2bx +c

find the roots of this function (critical points).

f '(0) =0 ----> c =0
f'(4) = 3a(16) +8b =0
48a +8b =0
b = -6a
f(x) = ax^3 -6ax^2 +d
f(0) =3 ----> d =3
f(4) =12 ----> 64a -96a +3 = 12
-32a = 9
a =-9/32 ----> b =-6a = 54/32 = 27/16

I found f(x) = -9x^3/32 +27x^2/16 +3

Answer 3

f(x) = ax^3+bx^2+cx+d
f '(x) = 3ax^2 + 2bx + c
12 = 48a +8b+c
3 = 0+0+c --> c=0
d= 3
12 = 64a+16b +4c+3
64a +16b =9
48a +8b = 12
-96a -16b =-24
-32a= -15
a = 15/32
b = (12-48(15/32))/8 = -1.3125
The equation is 15x^3/32 -1.3125x^2 +3 = 0
made an error somewhere