Calculus/Calc Help.

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Expert: - 11/29/2009


Question
Let f be a differentiable function with the following properties
a. f'(x)=ax^2+bx
b. f'(1)=-6 and f"(1)=6
c. The interval [1,2] f(x)dx=14

find f(x)


Please explain !

Answer
a. Given that f'(x) = ax² + bx, it is known that f(x) = ax³/3 + bx²/2 + C.

b. Given that f'(1) = -6, and that f'(x) = ax² + bx, that says -6 = a + b.
We know that f"(x) = the derivative of f'(x), so f"(x) = 2ax + b.
Since we are given that f"(1)= 6, we know that 6 = 2a + b.
Subtracting -6 = a + b from 6 = 2a + b, we get 12 = a.
Put 12 into both equations and on the 1st one get -6 = 12 + b => b = -18.
Put 12 into the 2nd one and get 6 = 24 - 18, and that makes sure the -18 was correct.

So far, we have f(x) = 12x³/3 - 18x²/2 + C = 4x³ - 9x² + C.

c.  for c., is that ∫f(x) dx from 1 to 2?
If so, the integral is x^4 - 3x³ + Cx, from 2 downto 1.
Putting these values in gives us (16-24+2C)-(1-3+C) = -8+2C + 2-C = 14.
So, C - 6 = 14, and C = 20.

Answer
This makes the function f(x) = 4x³ - 9x² + 20.

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