You need to use the inverse of differentiation, hence you need to integrate the given function twice to find the original function f(x).

`int f''(x) dx = f'(x) + c`

`int (6 + 6x + 24x^2)dx = int 6dx + int 6x dx + int 24x^2 dx`

`int (6 + 6x + 24x^2)dx = 6x + 6x^2/2 + 24x^3/3 + c`

`int (6 + 6x + 24x^2)dx = 6x + 3x^2 + 8x^3 + c`

Supposing that f'(0)=5, you need to substitute 0 in f'(x) to find the constant term c such that:

`f'(0) = 0 + 0 + 0 + c`

`5 = c`

`f'(x) = 6x + 3x^2 + 8x^3 + 5`

You need to integrate f'(x) to find f(x) such that:

`int(6x + 3x^2 + 8x^3 + 5)dx = int 6x dx + int 3x^2dx + int 8x^3dx + int 5`

`int (6x + 3x^2 + 8x^3 + 5)dx = 6x^2/2 + 3x^3/3 + 8x^4/4 + 5x + c`

`int (6x + 3x^2 + 8x^3 + 5)dx = 3x^2 + x^3 + 2x^4 + 5x + c`

Supposing that f(1)=14, you need to substitute 1 in f(x) to find the constant term c such that:

`14 = 3 + 1 + 2 + 5 + c`

`c = 14 - 11`

`c =3`

**Hence, evaluating the function f(x) using integration under given conditions yields `f(x) = 3x^2 + x^3 + 2x^4 + 5x + 3.` **

integrate the f''(x) to get f'(x)

f''(x)=6+6x+(24x^2)

f'(x)=6x+3x^2 + 8x^3 + C

and i think you made a mistake because they should have given you a solution for the f'(x). Maybe you meant to write that f'(0)=5 and f(1)=14?

in that case...

f'(0)=6(0)+3(0^2)+8(0^3)+C=5

C=5

then:

f'(x)=6x+3x^2 + 8x^3 + 5

f(x)=3x^2 + x^3 + 2x^4 + 5x + C

f(1)=3(1)^2 + (1)^3 + 2(1)^4 + 5(1) + C=14

C=3

f(x)=3x^2 + x^3 + 2x^4 + 5x + 3