The derivative of a function gives f’(x) = ax^2 – (a+b)x +c. What are the values of a,b and c if f(0)= 2, f(4) = 8 and f(10) = 12.

The derivative of f(x) is  f’(x) = ax^2 – (a+b)x
+c

Now the integral of f'(x) is
f(x).

Int[ f'x)] = Int [ (ax^2 – (a+b)x +c)  dx
]

=Int [ (ax^2 ] - Int [(a+b)x] +
Int[c]

= ax^3/3 - ((a+b)/2)x^2 + cx
+C

Now f(x) = ax^3/3 - ((a+b)/2)x^2 + cx
+C

f(0) = 0

=> f(0) =
a*0^3/3 - ((a+b)/2)*0^2 + c*0 +C

=> 2 =
C

f(4) = 8

=> a*4^3/3 -
((a+b)/2)*4^2 + c*4 +2 =8

=> (a/3)*64 - ((a+b)/2)*16
+ c*4 +2 =8

=> (a/3)*32 - ((a+b)/2)*8 + c*2 +1
=4

=> (a/3)*32 - ((a+b)/2)*8 + c*2 -3 =
0

f(10) = 12

=>
a*10^3/3 - ((a+b)/2)*10^2 + c*10 +2 =12

=>
(a/3)*1000 - ((a+b)/2)*100 + c*10 +2 =12

=>
(a/3)*500 - ((a+b)/2)*50 + c*5 +1 =6

=> (a/3)*500 -
((a+b)/2)*50 + c*5 -5 = 0

=> (a/3)*100 -
((a+b)/2)*10 + c - 1 = 0

Now we have two equations
:

(a/3)*32 - ((a+b)/2)*8 + c*2 -3 =
0

(a/3)*100 - ((a+b)/2)*10 + c - 1 =
0

and 3 variables a, b and
c.

So we cannot find unique values for all of
them.