Question

Find
f. f ''(x) = −2 + 36x − 12x2, f(0) = 2, f '(0) = 18

Answer 1

f''(x)=-2+36x-12x^2, f(0)=2, f'(0)=18
f'(x)=-2x+18x^2-4x^3+c
f'(0)=-2(0)+18(0)^2-4(0)^3+c=18
f'(0)=18x^2-4x^3-2x+18
f(x)=6x^3-x^4-x^2+18x+c
f(0)=6(0)^3-(0)^4-(0)^2+18(0)+c=2
f(x)=6x^3-x^4-x^2+18x+2

Answer 2

`f(x)=-x^4+6x^3-x^2+18x+2`

Explanation:

In order to find
`f(x)` we need to integrate twice
`f''(x)` . So it's good to have in mind the next rules:

`\int\ {x^n} \, dx =(1)/(n+1) x^(n+1) +C`

`\int\ {k} \, dx =k*x+C`

`k\in R`

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:

`\int\ {f(x)+g(x)} \, dx = \int\ {f(x)} \, dx + \int\ {g(x)} \, dx`

So, let's integrate
`f''(x)` in order to obtain
`f'(x)` :

`f'(x)= \int\ {f''(x)} \, dx =\int\ {-2+36x-12x^2} \, dx = \int\ {-2} dx+\int\ {36x} \, dx+\int\ {-12x^2} \, dx`

Integrating:

`\int\ {-2} dx+\int\ {36x} \, dx+\int\ {-12x^2} \, dx=-2x+(36)/(2) x^2-(12)/(3)x^3+C_1 \\=-2x+18x^2-4x^3+C_1`

Evaluating the initial condition in order to find C1:

`f'(0)=-2(0)+18(0)^2-4(0)^3+C_1=18\\C_1=18`

Now, let's integrate
`f'(x)` in order to obtain
`f(x)` :

`f(x)=\int\ {f'(x)} \, dx = \int\ {-2x+18x^2-4x^3+18} \, dx =(-2)/(2)x^2 +(18)/(3)x^3-(4)/(4) x^4+18x+C_2\\ f(x)=-x^2+6x^3-x^4+18x+C_2`

Evaluating the other initial condition in order to find C2:

`f(0)=-(0)^2+6(0)^3-(0)^4+18(0)+C_2=2\\C_2=2`

Knowing the value of C1 and C2, we can conclude that the function
`f(x)` is given by:

`f(x)=-x^4+6x^3-x^2+18x+2`