vinaumase05
Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

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

Sagot :

LammettHash

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

We appreciate your time. Please come back anytime for the latest information and answers to your questions. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.