f′′(x) = 1x−1 and f′(−2) = 6 and f(−2) = −6. Find f′(x) and find f(2)

Answer 1

f′′(x) = 1x−1 and f′(−2) = 6 and f(−2) = −6. Find f′(x) and find f(2)

first of all ∫f′′(x) dx =∫ ( 1x−1)dx

f'(x) = 1/2 x^2 -x + c

one more time take Integral of both side

∫f'(x)dx =∫ ( 1/2 x^2 -x + c) dx

f(x) = 1/6 x^3 - 1/2 x^2 + cx + D

now use of given information such as f′(−2) = 6 --------> 1/2 (-2)^2 -(-2) +c = 6 ------> 2 +2 +c = 6 -----> c=2

another given information is f(−2) = −6 ------> 1/6 (-2)^3 -1/2(-2)^2 + 2(-2) + D = -6

-8/3 -4/2 -4 + D = -6 -------> D = 8/3

Now you know f(x) = 1/6 x^3 - 1/2 x^2 + 2x + 8/3

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for finding f(2) , it is enough replace 2 to the f(x)

f(2) = 1/6 (2)^3 -1/2(2)^2 + 2.2 + 8/3

I hope it is useful,

Minoo