Critical points of a function and discontinuity
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I just wanted to ask, why does the following function:
$$f(x)=x^{1/3}(x+3)^{2/3}$$
Have 3 crtical points $0,-1,-3$, because its first derivative is:
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}$$
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}=0$$
Then $x=-1$ is the critical point as its undefined on $-3$, and $0$, so why is $-3$, and $0$ also considered a critical point$?$
derivatives
asked Nov 25 at 14:48
Aurora Borealis
833414
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up vote
0
down vote
favorite
I just wanted to ask, why does the following function:
$$f(x)=x^{1/3}(x+3)^{2/3}$$
Have 3 crtical points $0,-1,-3$, because its first derivative is:
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}$$
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}=0$$
Then $x=-1$ is the critical point as its undefined on $-3$, and $0$, so why is $-3$, and $0$ also considered a critical point$?$
derivatives
asked Nov 25 at 14:48
Aurora Borealis
833414
@caverac Why need $x ge 0$? Those are cuberoots in the formula.
– coffeemath
Nov 25 at 15:01
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I just wanted to ask, why does the following function:
$$f(x)=x^{1/3}(x+3)^{2/3}$$
Have 3 crtical points $0,-1,-3$, because its first derivative is:
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}$$
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}=0$$
Then $x=-1$ is the critical point as its undefined on $-3$, and $0$, so why is $-3$, and $0$ also considered a critical point$?$
derivatives
asked Nov 25 at 14:48
Aurora Borealis
833414
I just wanted to ask, why does the following function:
$$f(x)=x^{1/3}(x+3)^{2/3}$$
Have 3 crtical points $0,-1,-3$, because its first derivative is:
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}$$
$$f'(x)=frac{x+1}{x^{2/3}(x+3)^{1/3}}=0$$
Then $x=-1$ is the critical point as its undefined on $-3$, and $0$, so why is $-3$, and $0$ also considered a critical point$?$
derivatives
derivatives
asked Nov 25 at 14:48
Aurora Borealis
833414
asked Nov 25 at 14:48
Aurora Borealis
833414
edited Nov 25 at 16:20
asked Nov 25 at 14:48
Aurora Borealis
833414
asked Nov 25 at 14:48
Aurora Borealis
833414
asked Nov 25 at 14:48
Aurora Borealis
833414
833414
@caverac Why need $x ge 0$? Those are cuberoots in the formula.
– coffeemath
Nov 25 at 15:01
add a comment |
@caverac Why need $x ge 0$? Those are cuberoots in the formula.
– coffeemath
Nov 25 at 15:01
@caverac Why need $x ge 0$? Those are cuberoots in the formula.
– coffeemath
Nov 25 at 15:01
@caverac Why need $x ge 0$? Those are cuberoots in the formula.
– coffeemath
Nov 25 at 15:01
add a comment |
2 Answers
2
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oldest
votes
up vote
1
down vote
accepted
A critical point $c$ is where $f(c)$ defined, and $f'(c)$ either zero or undefined.
answered Nov 25 at 14:57
coffeemath
2,0851413
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up vote
0
down vote
Hint critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined.
or
a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.
answered Nov 25 at 14:56
John Nash
6818
1
Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition.
– coffeemath
Nov 25 at 14:58
some authors mentioned this point
– John Nash
Nov 25 at 15:00
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function.
– coffeemath
Nov 25 at 15:03
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
A critical point $c$ is where $f(c)$ defined, and $f'(c)$ either zero or undefined.
answered Nov 25 at 14:57
coffeemath
2,0851413
add a comment |
up vote
1
down vote
accepted
A critical point $c$ is where $f(c)$ defined, and $f'(c)$ either zero or undefined.
answered Nov 25 at 14:57
coffeemath
2,0851413
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
A critical point $c$ is where $f(c)$ defined, and $f'(c)$ either zero or undefined.
answered Nov 25 at 14:57
coffeemath
2,0851413
A critical point $c$ is where $f(c)$ defined, and $f'(c)$ either zero or undefined.
answered Nov 25 at 14:57
coffeemath
2,0851413
answered Nov 25 at 14:57
coffeemath
2,0851413
answered Nov 25 at 14:57
coffeemath
2,0851413
answered Nov 25 at 14:57
coffeemath
2,0851413
2,0851413
add a comment |
add a comment |
up vote
0
down vote
Hint critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined.
or
a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.
answered Nov 25 at 14:56
John Nash
6818
1
Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition.
– coffeemath
Nov 25 at 14:58
some authors mentioned this point
– John Nash
Nov 25 at 15:00
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function.
– coffeemath
Nov 25 at 15:03
add a comment |
up vote
0
down vote
Hint critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined.
or
a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.
answered Nov 25 at 14:56
John Nash
6818
1
Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition.
– coffeemath
Nov 25 at 14:58
some authors mentioned this point
– John Nash
Nov 25 at 15:00
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function.
– coffeemath
Nov 25 at 15:03
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined.
or
a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.
answered Nov 25 at 14:56
John Nash
6818
Hint critical points any limit points where the function may be prolongated by continuity or where the derivative is not defined.
or
a critical point or stationary point of a differentiable function of a real or complex variable is any value in its domain where its derivative is 0.
answered Nov 25 at 14:56
John Nash
6818
edited Nov 25 at 14:59
answered Nov 25 at 14:56
John Nash
6818
answered Nov 25 at 14:56
John Nash
6818
answered Nov 25 at 14:56
John Nash
6818
6818
1
Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition.
– coffeemath
Nov 25 at 14:58
some authors mentioned this point
– John Nash
Nov 25 at 15:00
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function.
– coffeemath
Nov 25 at 15:03
add a comment |
1
Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition.
– coffeemath
Nov 25 at 14:58
some authors mentioned this point
– John Nash
Nov 25 at 15:00
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function.
– coffeemath
Nov 25 at 15:03
1
1
Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition.
– coffeemath
Nov 25 at 14:58
Note that the function needs to be defined at the critical point. No "prolongation by continuity" is used in the definition.
– coffeemath
Nov 25 at 14:58
some authors mentioned this point
– John Nash
Nov 25 at 15:00
some authors mentioned this point
– John Nash
Nov 25 at 15:00
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function.
– coffeemath
Nov 25 at 15:03
John-- I'd say if one prolongs a domain by conytinuity one is working with a different function.
– coffeemath
Nov 25 at 15:03
add a comment |
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@caverac Why need $x ge 0$? Those are cuberoots in the formula.
– coffeemath
Nov 25 at 15:01