f ″(x) = x−2, x > 0, f(1) = 0, f(5) = 0

Question

f ″(x) = x−2, x > 0, f(1) = 0, f(5) = 0

find f(x)

Answer 1

f ″(x) = x − 2, x > 0, f(1) = 0, f(5) = 0. find f(x) = ??

Detailed Solution:

Given/Known: f ^″(x) = x − 2, f(1) = 0, f(5) = 0. f(x) = ??

f ^' (x) = ∫ f ^″(x) dx = ∫ (x − 2) dx = x²/2 − 2x + C

f(x) = ∫ f ^'(x) dx = ∫ (x²/2 − 2x + C) dx = x³/6 − x² + Cx + K

f(x) = x³/6 − x² + Cx + K = (x)³/6 − (x)² + C(x) + K

f(1) = 0==>f(1) = (1)³/6 − (1)² + C(1) + K = 0 ==>1/6 − 1 + C + K = 0==> C + K = 5/6 ==> Eqn1

f(5) = 0==>f(5) = (5)³/6 − (5)² + C(5) + K = 0==>125/6 − 25 + 5C + K = 0==>5C + K = 25/6 ==>Eqn2

C + K = 5/6 ==> Equation 1

5C + K = 25/6 ==> Equation 2

Equation 2 − Equation 1 ==> (5C + K = 25/6) − (C + K = 5/6) ==> 4C = 25/6 − 5/6 ==> 4C = 20/6

C = 5/6

5/6 + K = 5/6 <== back substitute into Equation 1

K = 0

f(x) = x³/6 − x² + Cx + K = x³/6 − x² + 5x/6 + 0 = x³/6 − x² + 5x/6

f(x) = x³/6 − x² + 5x/6 <== Final Answer

Check:

f(x) = x³/6 − x² + 5x/6

f ^' (x) = 3x²/6 − 2x+ 5/6 = x²/2 − 2x+ 5/6 (1st Derivative)

f ^″(x) = 2x/2 − 2 + 0 = x − 2 (2st Derivative)

Answer 2

Hello Fatima,

You would need to take the antiderivative of f"(x) twice

f'(x) = (1/2)*x² - 2x + C

f(x) = (1/6)*x³ - x² + Cx + D

To find C and D, use f(1) = 0 and f(5) = 0 to create 2 equations

f(1) = 0 ---> (1/6) - 1 + C + D = 0 -----> -5/6 + C + D = 0 ----> C + D = 5/6

f(5) = 0 ----> (1/6)*(125) - 25 + 5C + D = 0 -----> (125/6) -25 + 5C + D = 0 ----> -25/6 + 5C + D =0 ----> 5C+D=25/6

So your 2 equations are:

C + D = 5/6

5C+ D = 25/6

You subtract both equations to eliminate D: -4C = -20/6 ----> C = 20/24 ---> C = 5/6

Now insert 5/6 in one of the 2 equations. Let's do the 1st one: 5/6 + D = 0 ---> D = 0

so f(x) = (1/6)*x³ - x² + (5/6)*x

Checking for accuracy: f(1) = (1/6)*1 - 1 + 5/6 = -5/6 + 5/6 = 0

f(5) = (1/6)*5³ - 5² + (5/6)*5 = 125/6 - 25 + 25/6 = -25/6 + 25/6 = 0

f'(x) = (1/6)*3x² - 2x + 5/6

f"(x) = (1/6)*3*2x -2 = (1/6)*6x - 2 = x - 2

Check successful, the answer is correct.