Question

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

Answer 1

Answer:

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

Step-by-step 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:

$\int\ {x^n} \, dx =\frac{1}{n+1} x^{n+1} +C$

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

$k\in R$

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:

$\int\ {-2} dx+\int\ {36x} \, dx+\int\ {-12x^2} \, dx=-2x+\frac{36}{2} x^2-\frac{12}{3}x^3+C_1 \\=-2x+18x^2-4x^3+C_1$

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 $f(x)$ :

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

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:

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

Answer 2

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