The curve y=2x^3+ax^2+bx-30 has a stationary point when x=3. The curve passes through the point (4,2). (A) Find the value of a and the value of b.

#secondderivative #stationarypoints

A stationary point at x = 3 means the derivative dy/dx = 0 at that point. Differentiating, we have

dy/dx = 6x ² + 2ax + b

and so when x = 3,

0 = 54 + 6a + b

or

6a + b = -54 … … … [eq1]

The curve passes through the point (4, 2), which is to say y = 2 when x = 4. So we also have

2 = 128 + 16a + 4b - 30

or

16a + 4b = -96

4a + b = -24 … … … [eq2]

Eliminate b by subtracting [eq2] from [eq1] and solve for a, then for b :

(6a + b) - (4a + b) = -54 - (-24)

2a = -30

a = -15   ===>   b = 96

17.08

correct me if i am wrong.