Find f for the function.

Question

f ''(x) = 4 − 6x − 60x³,    f(0) = 3,    f '(0) = 6

Answer 1

Solution:

f'(x) = ∫ f''(x) = ∫ (4-6x-60 x³) dx = 4x - 6(x²/2) -60( x⁴/4) + C1 = 4x - 3x² - 15x⁴ + C1

but f'(0) = 6 , and 6 = 4(0) -3(0)² -15 (0)⁴ + C1, C1 =6

f'(x) = 4x - 3x²- 15x⁴ + 6

f(x) = ∫ f'(x) = ∫( 4x - 3x² - 15x⁴ + 6) dx = 4(x²/2) -3( x³/3) -15 (x⁵/5) + 6x + C2

f(x) = 2x² - x³ - 3x⁵ + 6x + C2

but f(0) = 3 = 2(0)² -(0)³ -3(0)⁵ + 6(0) + C2, C2 =3

f(x) = 2x² - x³ - 3x⁵ + 6x + 3

Answer 2

If f ''(x) = 4 - 6x - 60x³ then think of a function that will give you "4 - 6x - 60x³" when you take the derivative of it, and that will be f '(x).

So f '(x) = 4x - 3x² - 15x⁴ + any constant term (since taking the derivative of any constant is zero.

Since f '(0) = 6, we can solve for the constant:

0 = 4(0) - 3(0)² - 15(0)⁴ + C therefore C = 6 meaning f '(x) = 4x - 3x² - 15x⁴ + 6

Now repeat the process to get f(x):

f(x) = 2x² - x³ - 3x⁵ + 6x + C and since f(0) = 3:

3 = 2(0)² - (0)³ - 3(0)⁵ + 6(0) + C meaning C = 3