Finding the Minimum of a Function for Calculus Students

[BilloRani2012]

Homework Statement

Check for minimum:
When you've got your x value and sub it back into the f'(x) equation, should you get zero if it's a minimum?

Homework Equations

The Attempt at a Solution

 

[Disconnected]

Homework Statement

Check for minimum:
When you've got your x value and sub it back into the f'(x) equation, should you get zero if it's a minimum?

Homework Equations

The Attempt at a Solution

Absolutely!
If you got some value other then zero for the rate of change, then the value just to one side of that point would have a lower value, right? So it wouldn't be a minimum!

 

[BilloRani2012]

okay thanks :)

could you please help me with this question:

Any two vectors that are not parallel define a plane. So p = i + j - k and q = 2i + j define a plane. For what values of x is the vector r = xi + j + k in this plane?

ITS DUE TMRW!

Thanks :)

 

[Disconnected]

okay thanks :)

could you please help me with this question:

Any two vectors that are not parallel define a plane. So p = i + j - k and q = 2i + j define a plane. For what values of x is the vector r = xi + j + k in this plane?

ITS DUE TMRW!

Thanks :)

Sure, I'll help. But only if you give it a crack first and tell me where you run into trouble.

 

[Ray Vickson]

Absolutely!
If you got some value other then zero for the rate of change, then the value just to one side of that point would have a lower value, right? So it wouldn't be a minimum!

What is the minimum of f(x) = x on the interval 1 <= x <= 2? Is the derivative of f equal to zero there?

RGV

 

[Mark44]

Homework Statement

Check for minimum:
When you've got your x value and sub it back into the f'(x) equation, should you get zero if it's a minimum?

Aren't you going backwards here? Presumably you got an equation by setting f'(x) to zero, and then you solved for x in the equation f'(x) = 0. The solutions to this equation are possible candidates for being minima or maxima or neither.

Some examples:
1) f(x) = x², f'(x) = 2x and f'(x) = 0 for x = 0. There is a (global) minimum at (0, 0).
2) g(x) = x³, g'(x) = 3x² and g'(x) = 0 for x = 0. It turns out that this function has neither a minimum or maximum value of any kind (local or global).
3) h(x) = |x|, h'(x) = 1 if x > 0 and h'(x) = -1 if x < 0. There is a global minimum at (0, 0) even though there is no value of x for which h'(x) = 0.
4) (Ray's example) f(x) = x on [1, 2], f'(x) = 1. There is a minimum at (1, 1) and a maximum at (2, 2), even though f'(x) is never 0.

 

[Disconnected]

What is the minimum of f(x) = x on the interval 1 <= x <= 2? Is the derivative of f equal to zero there?

RGV

Of course. Very good point that I missed completely. I was thinking global minimums.

 

[Mark44]

Of course. Very good point that I missed completely. I was thinking global minimums.

If you goal is finding global minima or maxima, you want to look at
1) values of x for which f'(x) = 0.
2) values of x in the domain of f for which f' is undefined.
3) endpoints of an interval on which the function is defined.

 

[BilloRani2012]

Okay so the question was:

Any two vectors that are not parallel define a plane. So p = i + j - k and q = 2i + j define a plane. For what values of x is the vector r = xi + j + k in this plane?

My tutor said to find the the dot product of p and q. But we can't because p has 3 values and q just has 2 values??

 

[SammyS]

q = 2i + j +0k

That gives you the third component.

 

[BilloRani2012]

okay. so do i just find the dot product of p and q? But then how would i find x??

 

[SammyS]

Okay so the question was:

Any two vectors that are not parallel define a plane. So p = i + j - k and q = 2i + j define a plane. For what values of x is the vector r = xi + j + k in this plane?

My tutor said to find the the dot product of p and q. But we can't because p has 3 values and q just has 2 values??

First of all: I don't see what this has to do with the Original Post in this thread -- the question about the minimum.

For the question regarding the vectors:

I suggest finding the cross product (vector product), p × q. That vector is normal to the plane determined by p & q. Then for r to be in that same plane, it must also be perpendicular to the vector p × q .

 

[Mark44]

To echo what LCKurtz said, you should start a new thread when you have a new problem.

Here's a different approach to your vector problem. Since r is in the plane that is defined by p and q, r has to be a linear combination of p and q. By linear combination, I mean a sum of scalar multiples of p and q. This means that r = c₁p + c₂q, for some scalars (real constants) c₁ and c₂.

Two vectors are equal if and only if their corresponding components are equal. If you set the x, y, and z components equal in the equation above, you will get three equations in three unknowns, which you can solve for c₁, c₂, and x.

 

[BilloRani2012]

TO Mark44: So do u mean set x = c1p + c2p, y = c1p + c2p and z = c1p + c2p

 

[Mark44]

No. Set the x coordinate of r equal to the x coordinate of c₁p + c2q, set the y coordinate of r equal to the y coordinate of c₁p + c2q, and set the z coordinate of r equal to the z coordinate of c₁p + c2q. Maybe that's what you meant, but that isn't what you said.

 

[Disconnected]

If you goal is finding global minima or maxima, you want to look at
1) values of x for which f'(x) = 0.
2) values of x in the domain of f for which f' is undefined.
3) endpoints of an interval on which the function is defined.

[STRIKE]I hate everybody.[/STRIKE] It's entirely possible that I should just stop [STRIKE]trying to help[/STRIKE] confrusing people with wrong advice.

 

[HallsofIvy]

This same problem was posted under "physics homework" so I am merging this with that thread.