Derivative of a Point of Max: Finding a & b Conditions

[greg_rack]

Homework Statement

Given the function $y=ax^3+bx^2+2x-1$, find which values must $a$ and $b$ assume in order to have a relative maximum for $x=-1$.

Relevant Equations

none

The derivative of a point of maximum must be zero, and since
$$y'=3ax^2+2bx+2 \rightarrow y'(-1)=3a-2b+2 \rightarrow 3a-2b+2=0$$
we get the first condition for $a$ and $b$.

Now, since we want $x=-1$ to be a local maximum, the derivative of the function must be positive when tending to the left of $x=-1$, and negative when tending to the right of $x=-1$.
I believe this is the second point that would allow me to get $a$ and $b$... but I don't know how to "write it down" as a formal condition.

 

[Doc Al]

Hint: What does the second derivative tell you?

 

[greg_rack]

Yeah right, the second derivative must be negative at a local maximum!

PS: I take the opportunity to ask you why is that, since my textbook isn't really clear about it... I'm not sure about the geometrical/analytical "meaning" of the second derivative.

 

[pasmith]

If the derivative vanishes at x_0, then locally the function looks like f(x_0) + \frac12{f''(x_0)} (x-x_0)^2. By comparison to \pm x^2 you can see that this is a local maximum if f''(x_0) < 0 and a local minimum if f''(x_0) > 0.

 

[Doc Al]

Yeah right, the second derivative must be negative at a local maximum!

Exactly!

PS: I take the opportunity to ask you why is that, since my textbook isn't really clear about it... I'm not sure about the geometrical/analytical "meaning" of the second derivative.

You were heading right towards the answer in your thinking. The second derivative tells you how the first derivative (which is the slope of the original function) changes. Look up the "second derivative test" for finding local minima/maxima. Here's one explanation that might prove helpful: How to Use the Second Derivative Test

 

[epenguin]

Yeah right, the second derivative must be negative at a local maximum!

PS: I take the opportunity to ask you why is that, since my textbook isn't really clear about it... I'm not sure about the geometrical/analytical "meaning" of the second derivative.

It's the derivative of the derivative. And you want to know whether the slope as you pass through the extremum is increasing or decreasing.

 

[greg_rack]

Thanks guys, got it! @pasmith @Doc Al
Grazie @epenguin :)

 

[Office_Shredder]

Yeah I think you really already explained the test. You want the derivative to be positive to the left and negative to the right. That means it has to be decreasing. That means it's derivative must be negative (or potentially zero)