Question 702267
f '' (x) = 3x^2 - 2


Integrate both sides to get


f ' (x) = x^3 - 2x + C


We're told that f ' (-1) = 5, so


f ' (x) = x^3 - 2x + C


f ' (-1) = (-1)^3 - 2(-1) + C


5 = -1 + 2 + C


5 = 1 + C


5 - 1 = C


4 = C


C = 4


So f ' (x) = x^3 - 2x + 4


Integrate both sides to get


f(x) = (1/4)*x^4 - x^2 + 4x + C


Now use the fact that f(2) = -1


f(x) = (1/4)*x^4 - x^2 + 4x + C


f(2) = (1/4)*(2)^4 - (2)^2 + 4(2) + C


f(2) = (1/4)*(16) - 4 + 4(2) + C


f(2) = 4 - 4 + 8 + C


f(2) = 8 + C


-1 = 8 + C


-1 - 8 = C


-9 = C


C = -9


Therefore, the function f(x) is {{{f(x) = expr(1/4)*x^4 - x^2 + 4x - 9}}}