Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = 4, f '(2) = 12, f(2) = 18

f(x) = 2*x^2 + 4*x + 2.

Step-by-step explanation:

Ok, we have that:

f''(x) = 4.

Then, integrating, we have that:

f'(x) = 4*x + a.

and from the conditions we know that:

f'(2) = 12 = 4*2 + a = 8 + a

12 = 8 + a

a = 12 - 8 = 4

then f'(x) = 4*x + 4.

now we can integrate again and get:

f(x) = (1/2)*4*x^2 + 4*x + b.

f(x) = 2*x^2 + 4x + b

and we know that f(2) = 18:

f(2) = 18 = 2*2^2 + 4*2 + b = 8 + 8 + b

18 = 16 + b

18 - 16 = b = 2.

Then the function is:

f(x) = 2*x^2 + 4*x + 2.

3 is greater then both of those numbers