Horizontal tangent problem

Question

Find a cubic polynomial that has horizontal tangents at the points (-3,5) and (3,-2).

Answer 1

Find a cubic polynomial that has horizontal tangents at the points (–3,5) and (3,–2).

The general formula for the cubic polynomial we're looking for is

f(x) = ax³ + bx² + cx + d

Since the slope of the cubic function will be zero at x = –3 and x = 3,

it's derivative f'(x) will be a quadratic function with zeroes at x = –3 and x = 3.

So we can write

f'(x) = 3ax² + 2bx + c = 3a(x + 3)(x – 3)

= 3a(x² – 9) = 3ax² – 27a

This means that

f(x) = ∫f'(x) dx = ∫(3ax² – 27a)dx = ax³ – 27ax + d

Now we just plug in (x, f(x)) = (–3, 5) and (x, f(x)) = (3, –2) into

f(x) = ax³ – 27ax + d

and use the resulting pair of equations to solve for coefficients a and d

f(–3) = a(–3)³ – 27a(–3) + d = 5 which means that –27a + 81a + d = 54a + d = 5

f(3) = a(3)³ – 27a(3) + d = –2 which means that 27a – 81a + d = –54a + d = –2

Adding the two equations eliminates a and gives

2d = 3 which means that d = 3/2

Subtracting the two equations eliminates d and gives

108a = 7 which means that a = 7/108

Plugging these values of a and d into f(x) = ax³ – 27ax + d gives

f(x) = (7/108)x³ – 27(7/108)x + 3/2

which simplifies to

f(x) = (7/108)x³ – (7/4)x + 3/2

Answer

7x³/108 – 7x/4 + 3/2

is the required cubic polynomial that has horizontal tangents at the points (-3,5) and (3,-2).