Find a cubic polynomial (that is, f ( x ) = a x 3 + b x 2 + c x + d ) that has horizontal tangents at the points (-2,2) and (2,-2).

Answer 1

f(x) = ax³ + bx² + cx + d

d/dx [f(x)] = 3ax² +2bx + c

f(-2) = a(-2)³ + b(-2)² + c(-2) + d = -8a + 4b - 2c + d = 2

f(2) = a(2)³ + b(2)² + c(2) + d = 8a + 4b + 2c + d = -2

f^/(-2) = 3a(-2)² + 2b(-2) + c = 12a - 4b + c = 0

f^/(2) = 3a(2)² + 2b(2) + c = 12a + 4b + c = 0

This is now a 4 by 4 system you have to solve (which is no longer calculus, but algebra).

Solving this system gives a = 1/8 b = 0 c = -3/2 d = 0

(Make sure you know how to solve this system, or else ask again on Wyzant). So the cubic polynomial is:

f(x) = (1/8)x³ + (0)x² + (-3/2)x + 0

OR

f(x) = (1/8)x³ - (3/2)x

Hope this helps!