Mixing
Derivative of max function
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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
asked Apr 21 '13 at 18:27
phil12phil12
5823816
$endgroup$
add a comment |
$begingroup$
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
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 |
$begingroup$
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
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
asked Apr 21 '13 at 18:27
phil12phil12
5823816
asked Apr 21 '13 at 18:27
phil12phil12
5823816
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
$begingroup$
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
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
$endgroup$
add a comment |
$begingroup$
HINT : Analyze this function on different intervals.
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 |
$begingroup$
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?
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
$begingroup$
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
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
$endgroup$
add a comment |
$begingroup$
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
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
$endgroup$
add a comment |
$begingroup$
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
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
$endgroup$
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
$$f'(x) = begin{cases} 2(x-1) & text{if } x leq 1\ 0 & text{if }x geq 1end{cases}$$
answered Apr 21 '13 at 18:32
user17762
add a comment |
add a comment |
$begingroup$
HINT : Analyze this function on different intervals.
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 |
$begingroup$
HINT : Analyze this function on different intervals.
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 |
$begingroup$
HINT : Analyze this function on different intervals.
answered Apr 21 '13 at 18:29
KalissarKalissar
18319
$endgroup$
HINT : Analyze this function on different intervals.
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
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 |
$begingroup$
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?
answered Jul 22 '17 at 6:14
khankhan
1488
$endgroup$
add a comment |
$begingroup$
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?
answered Jul 22 '17 at 6:14
khankhan
1488
$endgroup$
add a comment |
$begingroup$
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?
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?
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
answered Jul 22 '17 at 6:14
khankhan
1488
1488
add a comment |
add a comment |
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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