f '(x) = 3x^2 - 5x + c
f(2) = 3(2)^2 - 5(2) + c = 11 + c
Given
f'' (x)= 6 x − 5 and f' (−1) = −5 and f (−1) = 2
Find f '(x) =
and find f(2)=
2 answers
AAAaannndd the bot gets it wrong yet again!
f '(x) = 3x^2 - 5x + c
since f'(-1) = -5, we have
3+5+c = -5
c = -13 so
f'(x) = 3x^2 - 5x - 13
f(x) = x^3 - 5/2 x^2 - 13x + c
since f(-1) = 2, we have
-1 - 5/2 + 13 + c = 2
c = -15/2 so
f(x) = x^3 - 5/2 x^2 - 13x - 15/2
f(2) = 8 - 10 - 26 - 15/2 = -35.5
f '(x) = 3x^2 - 5x + c
since f'(-1) = -5, we have
3+5+c = -5
c = -13 so
f'(x) = 3x^2 - 5x - 13
f(x) = x^3 - 5/2 x^2 - 13x + c
since f(-1) = 2, we have
-1 - 5/2 + 13 + c = 2
c = -15/2 so
f(x) = x^3 - 5/2 x^2 - 13x - 15/2
f(2) = 8 - 10 - 26 - 15/2 = -35.5