# 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.

### 2 Answers | Add Yours

f'(x) =ax^2-(a+b)x+c.

Let the primitive function of f'(x) be f(x). Then f(x) = Int{ax^2-(a+b)x+c}dx.

f(x) = Int ax^3dx -Int( a+b)xdx +Int cdx

f(x)= (a/3)x^3-(1/2)(a+b)x^2 + cx + d ..........(1), as Int x^n dx = (x^n)/(n+1). d is the constant of integration.

f(0) = 2 =(a/4)*0-(1/2)*0+c*0+d , So d = 2.

f(4) = (1/3)a*4^3-(1/2)(a+b)*4^2+c*4 +2 = 2

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

Multiply by 3 :

64a -24a-24b +12c +6 = 24

40a -24b +12c = - 18

20a-12b +6c = -9.................(1).

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

Multiply by 6:

1000a-150a-150b +30c +6 = 36

850a -150b +30c = 30

85a-15b +3c = 3.....................(2).

Therefore there are a, b and c are 3 variables with 2 equations could be solved for a and b in terms of c.

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.**