Derivative of max function

21

$begingroup$

I was just wondering what the derivative of $f(x) = max(0,1-x)^{2}$ would be. What technique do you use to determine this derivative?

calculus

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edited Apr 21 '13 at 18:33

user17762

asked Apr 21 '13 at 18:27

phil12phil12

5823816

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Don't you mean $max(0, (1-x)^2)$?
  $endgroup$
  – Javier
  Apr 21 '13 at 19:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  since $(1-x)^2 ge 0 $ (for x a real number), $max(0,(1-x)^2)$ is equal to $(1-x)^2$
  $endgroup$
  – Andre Holzner
  Jun 8 '16 at 12:35
  
  
  

  
  
  
  

add a comment | 

21

$begingroup$

I was just wondering what the derivative of $f(x) = max(0,1-x)^{2}$ would be. What technique do you use to determine this derivative?

calculus

share|cite|improve this question

edited Apr 21 '13 at 18:33

user17762

asked Apr 21 '13 at 18:27

phil12phil12

5823816

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Don't you mean $max(0, (1-x)^2)$?
  $endgroup$
  – Javier
  Apr 21 '13 at 19:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  since $(1-x)^2 ge 0 $ (for x a real number), $max(0,(1-x)^2)$ is equal to $(1-x)^2$
  $endgroup$
  – Andre Holzner
  Jun 8 '16 at 12:35
  
  
  

  
  
  
  

add a comment | 

21

21

21

7

$begingroup$

I was just wondering what the derivative of $f(x) = max(0,1-x)^{2}$ would be. What technique do you use to determine this derivative?

calculus

share|cite|improve this question

edited Apr 21 '13 at 18:33

user17762

asked Apr 21 '13 at 18:27

phil12phil12

5823816

$endgroup$

I was just wondering what the derivative of $f(x) = max(0,1-x)^{2}$ would be. What technique do you use to determine this derivative?

calculus

calculus

share|cite|improve this question

edited Apr 21 '13 at 18:33

user17762

asked Apr 21 '13 at 18:27

phil12phil12

5823816

share|cite|improve this question

edited Apr 21 '13 at 18:33

user17762

asked Apr 21 '13 at 18:27

phil12phil12

5823816

share|cite|improve this question

share|cite|improve this question

edited Apr 21 '13 at 18:33

user17762

edited Apr 21 '13 at 18:33

user17762

edited Apr 21 '13 at 18:33

user17762

asked Apr 21 '13 at 18:27

phil12phil12

5823816

asked Apr 21 '13 at 18:27

phil12phil12

5823816

asked Apr 21 '13 at 18:27

phil12phil12

5823816

5823816

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Don't you mean $max(0, (1-x)^2)$?
  $endgroup$
  – Javier
  Apr 21 '13 at 19:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  since $(1-x)^2 ge 0 $ (for x a real number), $max(0,(1-x)^2)$ is equal to $(1-x)^2$
  $endgroup$
  – Andre Holzner
  Jun 8 '16 at 12:35
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Don't you mean $max(0, (1-x)^2)$?
  $endgroup$
  – Javier
  Apr 21 '13 at 19:05
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  since $(1-x)^2 ge 0 $ (for x a real number), $max(0,(1-x)^2)$ is equal to $(1-x)^2$
  $endgroup$
  – Andre Holzner
  Jun 8 '16 at 12:35
  
  
  

  
  
  
  

3

3

$begingroup$
Don't you mean $max(0, (1-x)^2)$?
$endgroup$
– Javier
Apr 21 '13 at 19:05

$begingroup$
Don't you mean $max(0, (1-x)^2)$?
$endgroup$
– Javier
Apr 21 '13 at 19:05

2

2

$begingroup$
since $(1-x)^2 ge 0 $ (for x a real number), $max(0,(1-x)^2)$ is equal to $(1-x)^2$
$endgroup$
– Andre Holzner
Jun 8 '16 at 12:35

$begingroup$
since $(1-x)^2 ge 0 $ (for x a real number), $max(0,(1-x)^2)$ is equal to $(1-x)^2$
$endgroup$
– Andre Holzner
Jun 8 '16 at 12:35

add a comment | 

3 Answers
3

active

oldest

votes

30

$begingroup$

It might be of help to sketch the function or write it without the $max$. We get
$$f(x) = begin{cases} (1-x)^2 & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
It is easy to work out the derivative everywhere except at $x=1$.
At $x=1$, work out explicitly from definition.
$$lim_{h to 0^+} dfrac{f(1+h) - f(1)}{h} = 0$$
$$lim_{h to 0^-} dfrac{f(1+h) - f(1)}{h} = lim_{hto 0^-} dfrac{h^2}{h} = 0$$
Hence, we have
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$

share|cite|improve this answer

answered Apr 21 '13 at 18:32

user17762

$endgroup$

add a comment | 

5

$begingroup$

HINT : Analyze this function on different intervals.

share|cite|improve this answer

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)?
  $endgroup$
  – phil12
  Apr 21 '13 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now what happens when x = 1 ? Does f'(1) exist ?
  $endgroup$
  – Kalissar
  Apr 21 '13 at 18:33
  

  
  
  
  

add a comment | 

2

$begingroup$

I would generalize this as follows:

Let

f(x) = max(0, g(x))^2

Then

f'(x) = [2*g(x)*g'(x)] if g(x) is > 0 else [0]

Does it make sense?

share|cite|improve this answer

answered Jul 22 '17 at 6:14

khankhan

1488

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

30

$begingroup$

It might be of help to sketch the function or write it without the $max$. We get
$$f(x) = begin{cases} (1-x)^2 & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
It is easy to work out the derivative everywhere except at $x=1$.
At $x=1$, work out explicitly from definition.
$$lim_{h to 0^+} dfrac{f(1+h) - f(1)}{h} = 0$$
$$lim_{h to 0^-} dfrac{f(1+h) - f(1)}{h} = lim_{hto 0^-} dfrac{h^2}{h} = 0$$
Hence, we have
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$

share|cite|improve this answer

answered Apr 21 '13 at 18:32

user17762

$endgroup$

add a comment | 

30

$begingroup$

It might be of help to sketch the function or write it without the $max$. We get
$$f(x) = begin{cases} (1-x)^2 & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
It is easy to work out the derivative everywhere except at $x=1$.
At $x=1$, work out explicitly from definition.
$$lim_{h to 0^+} dfrac{f(1+h) - f(1)}{h} = 0$$
$$lim_{h to 0^-} dfrac{f(1+h) - f(1)}{h} = lim_{hto 0^-} dfrac{h^2}{h} = 0$$
Hence, we have
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$

share|cite|improve this answer

answered Apr 21 '13 at 18:32

user17762

$endgroup$

add a comment | 

30

30

30

$begingroup$

It might be of help to sketch the function or write it without the $max$. We get
$$f(x) = begin{cases} (1-x)^2 & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
It is easy to work out the derivative everywhere except at $x=1$.
At $x=1$, work out explicitly from definition.
$$lim_{h to 0^+} dfrac{f(1+h) - f(1)}{h} = 0$$
$$lim_{h to 0^-} dfrac{f(1+h) - f(1)}{h} = lim_{hto 0^-} dfrac{h^2}{h} = 0$$
Hence, we have
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$

share|cite|improve this answer

answered Apr 21 '13 at 18:32

user17762

$endgroup$

It might be of help to sketch the function or write it without the $max$. We get
$$f(x) = begin{cases} (1-x)^2 & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
It is easy to work out the derivative everywhere except at $x=1$.
At $x=1$, work out explicitly from definition.
$$lim_{h to 0^+} dfrac{f(1+h) - f(1)}{h} = 0$$
$$lim_{h to 0^-} dfrac{f(1+h) - f(1)}{h} = lim_{hto 0^-} dfrac{h^2}{h} = 0$$
Hence, we have
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$

share|cite|improve this answer

answered Apr 21 '13 at 18:32

user17762

share|cite|improve this answer

share|cite|improve this answer

answered Apr 21 '13 at 18:32

user17762

answered Apr 21 '13 at 18:32

user17762

answered Apr 21 '13 at 18:32

user17762

add a comment | 

add a comment | 

5

$begingroup$

HINT : Analyze this function on different intervals.

share|cite|improve this answer

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)?
  $endgroup$
  – phil12
  Apr 21 '13 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now what happens when x = 1 ? Does f'(1) exist ?
  $endgroup$
  – Kalissar
  Apr 21 '13 at 18:33
  

  
  
  
  

add a comment | 

5

$begingroup$

HINT : Analyze this function on different intervals.

share|cite|improve this answer

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)?
  $endgroup$
  – phil12
  Apr 21 '13 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now what happens when x = 1 ? Does f'(1) exist ?
  $endgroup$
  – Kalissar
  Apr 21 '13 at 18:33
  

  
  
  
  

add a comment | 

5

5

5

$begingroup$

HINT : Analyze this function on different intervals.

share|cite|improve this answer

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

$endgroup$

HINT : Analyze this function on different intervals.

share|cite|improve this answer

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

share|cite|improve this answer

share|cite|improve this answer

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

answered Apr 21 '13 at 18:29

KalissarKalissar

18319

18319

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)?
  $endgroup$
  – phil12
  Apr 21 '13 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now what happens when x = 1 ? Does f'(1) exist ?
  $endgroup$
  – Kalissar
  Apr 21 '13 at 18:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)?
  $endgroup$
  – phil12
  Apr 21 '13 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now what happens when x = 1 ? Does f'(1) exist ?
  $endgroup$
  – Kalissar
  Apr 21 '13 at 18:33
  

  
  
  
  

$begingroup$
If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)?
$endgroup$
– phil12
Apr 21 '13 at 18:31

$begingroup$
If x >= 1, f'(x) = 0, else f'(x) = -2(1-x)?
$endgroup$
– phil12
Apr 21 '13 at 18:31

$begingroup$
Now what happens when x = 1 ? Does f'(1) exist ?
$endgroup$
– Kalissar
Apr 21 '13 at 18:33

$begingroup$
Now what happens when x = 1 ? Does f'(1) exist ?
$endgroup$
– Kalissar
Apr 21 '13 at 18:33

add a comment | 

2

$begingroup$

I would generalize this as follows:

Let

f(x) = max(0, g(x))^2

Then

f'(x) = [2*g(x)*g'(x)] if g(x) is > 0 else [0]

Does it make sense?

share|cite|improve this answer

answered Jul 22 '17 at 6:14

khankhan

1488

$endgroup$

add a comment | 

2

$begingroup$

I would generalize this as follows:

Let

f(x) = max(0, g(x))^2

Then

f'(x) = [2*g(x)*g'(x)] if g(x) is > 0 else [0]

Does it make sense?

share|cite|improve this answer

answered Jul 22 '17 at 6:14

khankhan

1488

$endgroup$

add a comment | 

2

2

2

$begingroup$

I would generalize this as follows:

Let

f(x) = max(0, g(x))^2

Then

f'(x) = [2*g(x)*g'(x)] if g(x) is > 0 else [0]

Does it make sense?

share|cite|improve this answer

answered Jul 22 '17 at 6:14

khankhan

1488

$endgroup$

I would generalize this as follows:

Let

f(x) = max(0, g(x))^2

Then

f'(x) = [2*g(x)*g'(x)] if g(x) is > 0 else [0]

Does it make sense?

share|cite|improve this answer

answered Jul 22 '17 at 6:14

khankhan

1488

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 '17 at 6:14

khankhan

1488

answered Jul 22 '17 at 6:14

khankhan

1488

answered Jul 22 '17 at 6:14

khankhan

1488

1488

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

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