Absolute equation to find the max and min point

[Lifeforbetter]

Homework Statement

|x -1| + |$x^2 $ - 2x|
Domain (0, 2)
Find the max and min point

Relevant Equations

F'(x) = 0
To find max and min point

X-1 > 0
x>1

$x^2$ - 2x >0
X>2

For 0<x<1
f(x) = -x +1 + -$x^2 $ +2x

For 1<x<2
f(x) = x -1 - $x^2 $ + 2x

Find each of the equation the critical point
By using f'(x) = 0
And decide which is max and min

I get x = 1/2 and x = 3/2
But it's wrong
Why?

 

[nuuskur]

This is what wolframalpha has to say on the matter.

Spoiler

On a pedantic note, fix your cases such that x=1 is considered in one of them (depends on which way you define absolute value).

You get two candidates for local extrema. Now check which kind they are, whether by definition or by the second derivative trick. What's the story with x=1?

As a reminder: if there is a local extremum at point a, then the derivative at that point is zero or the function is not differentiable at a (in other words, the point a is a critical point).

There is no equivalence, however. For x=1 there is no differentiability, so you will have to check by definition, whether it is a local extremum point.

 

[HallsofIvy]

The problem is to find maxima and minima for y= |x- 1|+ |x^2- 2x|= |x-1|+ |x(x- 2)|.

The first thing I would do is recognize that the absolute values are 0 for x= 1, x= 0, and x= 2. So I would divide the real line into four intervals: x< 0, 0< x< 1, 1< x< 2, and x> 2.

If x< 0 then all of x- 1, x, and x- 2 are negative. The product of x and x-2 however is positive so y= -(x- 1)+ x(x- 2)= -x+ 1+ x^2- 2x= x^2- 3x+ 1. y'= 2x- 3 which is 0 when x= 3/2. But that is not less than 0. y has no critical point when x< 0.

If 0< x< 1 then x is positive but x- 1 and x- 2 are negative. The product of x and x- 2 is negative so y= -(x-1)- x(x- 2)= -x+ 1- x^2+ 2x= -x^2+ x+ 1. y'= -2x+ 1 which is 0 when x= 1/2. Yes, that is between 0 and 1 so x= 1/2, y= -1/4+ 1/2+ 1= 5/4, is a critical point.

If 1< x< 2 then both x and x- 1 are positive but x- 2 is negative. The product of x and x- 2 is negative so y= x- 1- x(x- 2)= x- 1- x^2+ 2x= -x^2+ x- 1. y'= -2x+ 1 as before. That is 0 when x= 1/2 which is not between 1 and 2. There is no critical point between 1 and 2.

If x> 2 then all three of x, x- 1, and x- 2 are positive. So y= x- 1+ x(x- 2)= x- 1+ x^2- 2x= x^2- x- 1. y'= 2x- 1 which is 0 when x= 1/2. That is not larger than 2 so there is not critical point for x> 2.

The only critical point is x= 1/2, where y= 5/4 but we also need to check x= 0, x= 1, and x= 2. When x= 0, y= |0- 1|+ |0- 0|= |-1|= 1. When x= 1, y= |1- 1|+ |1- 2|= |-1|= 1. When x= 2, y= |2- 1|+ |4- 4|= |1|= 1.

The maximum value for the function is 5/4 which happens when x= 1/2. The minimum value is 1 which happens at x= 0, 1, and 2.

 

[SammyS]

;..

If 1< x< 2 then both x and x- 1 are positive but x- 2 is negative. The product of x and x- 2 is negative so y= x- 1- x(x- 2)= x- 1- x^2+ 2x= -x^2+ x- 1. y'= -2x+ 1 as before. That is 0 when x= 1/2 which is not between 1 and 2. There is no critical point between 1 and 2.
,,,

You have an error in the 1 < x < 2 case.
Should be $y=-x^2+3x-1$

 

[Lifeforbetter]

The problem is to find maxima and minima for y= |x- 1|+ |x^2- 2x|= |x-1|+ |x(x- 2)|.

The first thing I would do is recognize that the absolute values are 0 for x= 1, x= 0, and x= 2. So I would divide the real line into four intervals: x< 0, 0< x< 1, 1< x< 2, and x> 2.

If x< 0 then all of x- 1, x, and x- 2 are negative. The product of x and x-2 however is positive so y= -(x- 1)+ x(x- 2)= -x+ 1+ x^2- 2x= x^2- 3x+ 1. y'= 2x- 3 which is 0 when x= 3/2. But that is not less than 0. y has no critical point when x< 0.

If 0< x< 1 then x is positive but x- 1 and x- 2 are negative. The product of x and x- 2 is negative so y= -(x-1)- x(x- 2)= -x+ 1- x^2+ 2x= -x^2+ x+ 1. y'= -2x+ 1 which is 0 when x= 1/2. Yes, that is between 0 and 1 so x= 1/2, y= -1/4+ 1/2+ 1= 5/4, is a critical point.

If 1< x< 2 then both x and x- 1 are positive but x- 2 is negative. The product of x and x- 2 is negative so y= x- 1- x(x- 2)= x- 1- x^2+ 2x= -x^2+ x- 1. y'= -2x+ 1 as before. That is 0 when x= 1/2 which is not between 1 and 2. There is no critical point between 1 and 2.

If x> 2 then all three of x, x- 1, and x- 2 are positive. So y= x- 1+ x(x- 2)= x- 1+ x^2- 2x= x^2- x- 1. y'= 2x- 1 which is 0 when x= 1/2. That is not larger than 2 so there is not critical point for x> 2.

The only critical point is x= 1/2, where y= 5/4 but we also need to check x= 0, x= 1, and x= 2. When x= 0, y= |0- 1|+ |0- 0|= |-1|= 1. When x= 1, y= |1- 1|+ |1- 2|= |-1|= 1. When x= 2, y= |2- 1|+ |4- 4|= |1|= 1.

The maximum value for the function is 5/4 which happens when x= 1/2. The minimum value is 1 which happens at x= 0, 1, and 2.

Why 0 1 2 of interval included in the domain?

 

[HallsofIvy]

The original post said "Domain (0, 2)". 0 and 2 are not included but 1 is. So I should have said "the minimum value is 1 which happens at x= 1."

 

[epenguin]

I would say before any calculations I would do a sketch. In any old-fashioned school I think the teacher would do this in class. Graphing with devices most people have nowadays makes it less tedious too.

 

[SammyS]

I would say before any calculations I would do a sketch. In any old-fashioned school I think the teacher would do this in class. Graphing with devices most people have nowadays makes it less tedious too.

{Graph}

Just to note: @nuuskur had a graph from WolframAlpha in his Post (#2), but the penguin's graphs show how the overall graph emerges from its basic parts.

 

[epenguin]

Just to note: @nuuskur had a graph from WolframAlpha in his Post (#2), but the penguin's graphs show how the overall graph emerges from its basic parts.

Thank you, because it was behind a spoiler I hadn't noticed. I must also remember this device which I have never used I think.

 

[SammyS]

Thank you, because it was behind a spoiler I hadn't noticed. I must also remember this device which I have never used I think.

Yes.
I seldom use it either.

I suppose it's useful at times, but in this case it did more to "spoil" than to help.

 

[Merlin3189]

I still get the same maxima as Lifeforbetter's first post. Has anyone said yet why or how that is wrong?

I get x = 1/2 and x = 3/2
But it's wrong
Why?

 

[Lifeforbetter]

I still get the same maxima as Lifeforbetter's first post. Has anyone said yet why or how that is wrong?

That's it the critical points x. But the question asked for value which is y

 

[Mark44]

That's it the critical points x. But the question asked for value which is y

Just evaluate f at those critical numbers, with $f(x) = |x -1| + |x^2 - 2x|$. How hard is that?
If x = 1/2 is a critical number, then (1/2, f(1/2)) is a critical point, and f(1/2) would be the max. or min. value. The same holds for x = 3/2, assuming that 1/2 and 3/2 are the correct values.