The lines x + y = 0 and x + y = 4/3 are tangent to the graph

the lines x + y = 0 and x + y = 4/3 are tangent to the graph of y = F(x), where F is an antiderivative of -x^2. Find F

f ' (x) = -x^2

F(x) = -(1/3)x^3 + C

eqn of tangent is given by

y = F(a) + f ' (a)(x - a) where a is x- coordinate of point of tangency

since x + y = 0 and x + y = 4/3 are tangents, slope of tangent = -1 ==> f '(a) = -1

since f'(x) = -x^2 ==>f ' (a) = -a^2 ==> a^2 = 1 and a = -1 and 1

y = -(1/3) + C - 1(x - 1) and

y = -(1/3)(-1) + C - 1(x + 1)

=> y = -x + C + 2/3 and

y = -x + C - 2/3

x + y = C + 2/3

x + y = C - 2/3

==> C = 2/3

so F(x) = -(1/3)x^3 + 2/3

= 1/3[- x^3 + 2 ]




y = -x^3 + ax^2 + C - a^3/3