Finding Global Max and Min values of an Absolute Function

[Wi_N]

Homework Statement

y=0.5x^3 - abs(1-3x), [-2,2]

The Attempt at a Solution

So i found the min value y(-2)=-11 and i found lokal min at x=sqrt2
There is a sharp max somewhere around x=0.3 but i cannot for the life of me find how to get that point.
Also i wrote there is only 1 local min and that was wrong apparently but plotting it i only see one local min.

 

[Mark44]

Homework Statement

y=0.5x^3 - abs(1-3x), [-2,2]

The Attempt at a Solution

So i found the min value y(-2)=-11 and i found lokal min at x=sqrt2
There is a sharp max somewhere around x=0.3 but i cannot for the life of me find how to get that point.
Also i wrote there is only 1 local min and that was wrong apparently but plotting it i only see one local min.

Replace the absolute values by writing the function with a piecewise definition.
|1 - 3x| = 1 - 3x, if x <= 1/3, and |1 - 3x| = -(1 - 3x), if x >= 1/3. On each part of these intervals, you have a function whose graph is a straight line.

 

[Wi_N]

Replace the absolute values by writing the function with a piecewise definition.
|1 - 3x| = 1 - 3x, if x <= 1/3, and |1 - 3x| = -(1 - 3x), if x >= 1/3. On each part of these intervals, you have a function whose graph is a straight line.

1/54

thanks

 

[Mark44]

1/54

?
What does this have to do with your problem?

 

[Wi_N]

?
What does this have to do with your problem?

thats the max value of the function.

 

[Ray Vickson]

Homework Statement

y=0.5x^3 - abs(1-3x), [-2,2]

The Attempt at a Solution

So i found the min value y(-2)=-11 and i found lokal min at x=sqrt2
There is a sharp max somewhere around x=0.3 but i cannot for the life of me find how to get that point.
Also i wrote there is only 1 local min and that was wrong apparently but plotting it i only see one local min.

You cannot find the maximum by setting the derivative to zero, because the maximum occurs at a point of non-differeniability. In a case like this one, you can split up the function into two parts: (i) the part where $-2 \leq x \leq 1/3$; and (ii) the part where $1/3 \leq x \leq 2.$ On each part you can look for maxima and minima by finding interior optima (if any)----where the derivative vanishes---and checking for endpoint maxima or minima. Or, you can just plot the thing and inspect it visually.

In this case you have a global maximum at $x = 1/3$ and a local maximum at $x=2$. You have a global minimum at $x=-2$ and a local minimum at $x = \sqrt{2}.$ Note that at both global optima the derivative fails to vanish. The derivative is zero at the local minimum but nonzero at the local max.