Answer:
[tex]f(x)=-x^4+6x^3-x^2+18x+2[/tex]
Step-by-step explanation:
In order to find [tex]f(x)[/tex] we need to integrate twice [tex]f''(x)[/tex] . So it's good to have in mind the next rules:
[tex]\int\ {x^n} \, dx =\frac{1}{n+1} x^{n+1} +C[/tex]
[tex]\int\ {k} \, dx =k*x+C[/tex]
[tex]k\in R[/tex]
Where C, is an arbitrary constant.
The integral of the sum of two functions is equal to the sum of the integrals of these functions:
[tex]\int\ {f(x)+g(x)} \, dx = \int\ {f(x)} \, dx + \int\ {g(x)} \, dx[/tex]
So, let's integrate [tex]f''(x)[/tex] in order to obtain [tex]f'(x)[/tex] :
[tex]f'(x)= \int\ {f''(x)} \, dx =\int\ {-2+36x-12x^2} \, dx = \int\ {-2} dx+\int\ {36x} \, dx+\int\ {-12x^2} \, dx[/tex]
Integrating:
[tex]\int\ {-2} dx+\int\ {36x} \, dx+\int\ {-12x^2} \, dx=-2x+\frac{36}{2} x^2-\frac{12}{3}x^3+C_1 \\=-2x+18x^2-4x^3+C_1[/tex]
Evaluating the initial condition in order to find C1:
[tex]f'(0)=-2(0)+18(0)^2-4(0)^3+C_1=18\\C_1=18[/tex]
Now, let's integrate [tex]f'(x)[/tex] in order to obtain [tex]f(x)[/tex] :
[tex]f(x)=\int\ {f'(x)} \, dx = \int\ {-2x+18x^2-4x^3+18} \, dx =\frac{-2}{2}x^2 +\frac{18}{3}x^3-\frac{4}{4} x^4+18x+C_2\\ f(x)=-x^2+6x^3-x^4+18x+C_2[/tex]
Evaluating the other initial condition in order to find C2:
[tex]f(0)=-(0)^2+6(0)^3-(0)^4+18(0)+C_2=2\\C_2=2[/tex]
Knowing the value of C1 and C2, we can conclude that the function[tex]f(x)[/tex] is given by:
[tex]f(x)=-x^4+6x^3-x^2+18x+2[/tex]
