Differentiation, Quadratic Equations and the Discriminant

Question

Let f(x) = 5ax^3 - 2bx^2 +4cx

1 Find dy/dx f(x)

2 Given that f'(x) => 0 and a > 0, show that b^2 =< 15ac

Answer 1

For this question we need to assume that a, b, c are constants, as opposed functions of x.

Secondly I believe the notation dy/dx f(x) is meant to mean df/dx, or alternatively f'(x).

1) f'(x) = 15ax² - 4bx + 4c

2) Given that the parabola y = 15ax² - 4bx + 4c >= 0

the vertex, in particular, must have a non-negative y coordinate.

The x-coordinate of the vertex is x = 2b/15a.

(If you do not know why that is, see NOTE 1 below.)

Thus the y-coordinate of the vertex is

y = 15a(2b/15a)² - 4b(2b/15a) + 4c = 4b²/15a - 8b²/15a + 4c

We can simplify the inequality

4b²/15a - 8b²/15a + 4c >= 0 (i.e., "vertex must have non-negative y coordinate")

by multiplying it by 15a.

We can simply multiply it by 15a because of the given assumption a > 0.

4b² - 8b² + 60ac >= 0

After algebraic simplification

b² <= 15ac

NOTE 1: Derivation of the formula for vertex.

Given a general parabola

y = Ax² + Bx + C

the vertex is where the derivative is 0:

2Ax + B = 0

x = -B/2A

NOTE 2: The assumption a > 0 did not have to be given to you; it is implied by f'(x) >= 0.

15ax² - 4bx + 4c >= 0

implies that the coefficient of x² must be positive, i.e.,

15a > 0