Find the particular antiderivative that satisfies the following conditions:

Question

Find the particular antiderivative that satisfies the following conditions:

 

f''(x)=24x^3-6;   f(0)=1;   f'(0)=-3

 

f(x)=_____?

Answer 1

Hi Hussein.

 

The antiderivative is just "un-deriving".  So with a polynomial, we will ADD 1 to the exponent, then divide by the new exponent.

 

The antiderivative of f''(x) = 24x³ - 6  will be f'(x) = 24(x⁴/4) - 6(x¹/1) + C

 

Simplifying this, f'(x) = 6x⁴ - 6x + C  (to check, take the derivative and see that it equals 24x³ - 6)

To find C, we need the other information they gave us about f':  f'(0) = -3.   So plug x=0 into f'(x) 

 

f'(0) = 6(0)⁴ - 6(0) + C = -3  or  C = -3.   So now you have the full 1st derivative.  It is

 

f'(x) = 6x⁴ - 6x - 3

 

 

Now, un-derive  f'(x).

 

f(x) = 6(x⁵/5) - 6(x²/2) - 3(x¹/1) + K

 

Simplify this and get 

 

f(x) = (6/5)x⁵ - 3x² - 3x + K   (again, check by taking deriv and see if get f')

 

This time, we will find K using the additional info they provided on f(x).   f(0) = 1, so plug x=0 into f(x)

 

f(0) = (6/5)(0)⁵ - 3(0)² - 3(0) + K = 1  or K = 1

 

So the full, particular antiderivative is: 

 

f(x) = (6/5)x⁵ - 3x² - 3x + 1