Find the min and max value from absolute function?

[Helly123]

Homework Statement

Homework Equations

f ' (x) = 0 -> to find extreme point

The Attempt at a Solution

can I differentiate the function directly?
f ' (x) = |1| + | 2x - 2|
0 = 2x - 1
x = 1/2
and x from domain = { 0, 2 }

the range we get :
f(0) = 1
f(2) = 1
f(1/2) = 1/2 + 3/4 = 5/4

min = 1
max = 5/4

 

[SammyS]

Homework Statement

Homework Equations

f ' (x) = 0 -> to find extreme point

The Attempt at a Solution

can I differentiate the function directly?
f ' (x) = |1| + | 2x - 2|

That is not the correct derivative.

It may be helpful to graph each of the following:

$y=\left \vert x-1 \right \vert$
$y=\left \vert x^2-2x \right \vert$​

Furthermore:
There are more conditions for which you may have a minimum or maximum. What are the more general conditions?

 

[mjc123]

The answers are right, but the method is not. If f = |x-1|, f' is not |1|, as you can see by drawing the curve.
Let's simplify by noting that in the range 0-2, x²-2x is always negative, so |x²-2x| is 2x-x²
For |x-1|, consider the ranges 0-1 and 1-2 separately. (x=1/2 is not the only maximum.)

 

[Buffu]

Homework Statement

View attachment 206526

Homework Equations

f ' (x) = 0 -> to find extreme point

The Attempt at a Solution

can I differentiate the function directly?
f ' (x) = |1| + | 2x - 2|
0 = 2x - 1
x = 1/2
and x from domain = { 0, 2 }

the range we get :
f(0) = 1
f(2) = 1
f(1/2) = 1/2 + 3/4 = 5/4

min = 1
max = 5/4

Turn it into a composite function of three elementary functions, then do the derivative trick.

 

[Helly123]

$f(x) = | x-1 | + | x^2 -2x |$
f(x) = |x-1 |> 0
x > 1
For domain 0,2
Means for 0<= x < 1
x-1 negative
| x-1| = - (x-1) = 1-x

For 1<= x < = 2
x-1 positive

$f(x) = | x^2 - 2x |$
$f(x) = x^2 - 2x > 0$
x > 2

Means for domain {0,2}
0<= x <= 2
$f(x) = x^2 - 2x negative$
$f(x) = | x^2 - 2x | = -(x^2 -2x) = 2x - x^2$

For
0<= x < 1
$F(x) = 1-x + 2x - x^2$
Extreme point = -b/2a = 1/2, x = 1/2
For 1<= x <= 2
$F(x) = x - 1 + 2x - x^2$
Extreme point = -b/2a = 3/2, x = 3/2

Check for x = 0, x = 2, x = 1/2 , x = 3/2
F(0) = 1
F(1/2) = 3/2 - 1/4 = 5/4
F(2) = 1
F(3/2) = 1/2 + 3 - 9/4 = 2/4 + 12/4 - 9/4 = 5/4

Max = 5/4
Min = 1

 

[SammyS]

There may also be a min/max where the derivative is undefined.

 

[Helly123]

There may also be a min/max where the derivative is undefined.

what exactly is that?

 

[SammyS]

what exactly is that?

In your Opening Post, you stated:

Homework Equations

f ' (x) = 0 -> to find extreme point

It's also true that you may find an extreme point where the derivative, f ' (x), is undefined.

 

[Mark44]

There may also be a min/max where the derivative is undefined.

what exactly is that?

For example, g(x) = |x|.
g'(x) > 0 if x > 0, and g'(x) < 0 if x < 0.
This function has a minimum even though there is no point at which its derivative is zero.

 

[Helly123]

How to find max and min without differentiation ?

 

[Helly123]

Turn it into a composite function of three elementary functions, then do the derivative trick.

What do you meant "3" elementary functions? Aren't there only 2 ?

 

[Helly123]

The answers are right, but the method is not. If f = |x-1|, f' is not |1|, as you can see by drawing the curve.
Let's simplify by noting that in the range 0-2, x²-2x is always negative, so |x²-2x| is 2x-x²
For |x-1|, consider the ranges 0-1 and 1-2 separately. (x=1/2 is not the only maximum.)

To get rid of the absolute , we must add min if the function negative, and must add + if the function positive?

 

[Ray Vickson]

Homework Statement

View attachment 206526

Homework Equations

f ' (x) = 0 -> to find extreme point

No: banish forever from your mind the requirement that $f'(x) = 0$ for a max or a min on a finite interval $[a,b]$, as it is just not true. The correct statement is that "If differentiable $f(\cdot)$ has a maximum or a minimum at an interior point $a < x < b$ then $f'(x) = 0$." The derivative may not be zero at an end-point max or min. For example, the function $f(x) = x$ on the interval $[0,1]$ has minimum at $x = 0$ and a maximum at $x = 1$, but the derivative does not vanish at either of those two points (or, for that matter, at any point $x$).

 

[Buffu]

What do you meant "3" elementary functions? Aren't there only 2 ?

Why do you think there are only two ?

 

[Helly123]

Why do you think there are only two ?

Only |x -1| and |x^2 -2x| ?

 

[Mark44]

Turn it into a composite function of three elementary functions, then do the derivative trick.

It's not obvious to me that f(x) = |x| + |x² - 2x| can be viewed as the composition of three elementary functions. I think I get what you're hinting at, but a better word choice would be sum, rather than composition, which implies function composition.

In any case, this thread is posted in the Precalc section and the OP later asks how to answer the question without using derivatives. It's not clear what techniques should be used, though, as the OP started off by talking about derivatives.

@Helly123, this problem does not require the use of calculus. Sketch the graph of y = |x - 1|+ |x² - 2x|. This graph will have two formulas on the interval [0, 2]. It's possible to find the max. and min. values for each of these forumulas without the use of calculus.

 

[Buffu]

Only |x -1| and |x^2 -2x| ?

Consider your function between $(-\infty, 0], (0, 1], (1,2]$ and $(2, \infty)$.

 

[Ray Vickson]

Consider your function between $(-\infty, 0], (0, 1], (1,2]$ and $(2, \infty)$.

The question specified the domain $[0,2]$.