If $f(x)$ is odd, then prove that $f'(0)=0$
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If $f(x)$ is odd, then prove that $f'(0)=0$, assuming that $f'(0)$ exists.
We have that $f(x)$ is odd. Therefore, $f(-x)=-f(x)implies -f'(-x)=-f'(x)$. What I do after this?
functions derivatives even-and-odd-functions
edited Aug 15 at 4:15
Math Lover
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asked Aug 15 at 4:12
H.S.
6
add a comment |Â
up vote
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If $f(x)$ is odd, then prove that $f'(0)=0$, assuming that $f'(0)$ exists.
We have that $f(x)$ is odd. Therefore, $f(-x)=-f(x)implies -f'(-x)=-f'(x)$. What I do after this?
functions derivatives even-and-odd-functions
edited Aug 15 at 4:15
Math Lover
12.5k21232
asked Aug 15 at 4:12
H.S.
6
5
The result just isn't true. Maybe you wanted even $f$?
– Ian
Aug 15 at 4:19
1
$f(x) = x$ is odd, but $f'(x) = 1$.
– copper.hat
Aug 15 at 5:35
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $f(x)$ is odd, then prove that $f'(0)=0$, assuming that $f'(0)$ exists.
We have that $f(x)$ is odd. Therefore, $f(-x)=-f(x)implies -f'(-x)=-f'(x)$. What I do after this?
functions derivatives even-and-odd-functions
edited Aug 15 at 4:15
Math Lover
12.5k21232
asked Aug 15 at 4:12
H.S.
6
If $f(x)$ is odd, then prove that $f'(0)=0$, assuming that $f'(0)$ exists.
We have that $f(x)$ is odd. Therefore, $f(-x)=-f(x)implies -f'(-x)=-f'(x)$. What I do after this?
functions derivatives even-and-odd-functions
edited Aug 15 at 4:15
Math Lover
12.5k21232
asked Aug 15 at 4:12
H.S.
6
edited Aug 15 at 4:15
Math Lover
12.5k21232
edited Aug 15 at 4:15
Math Lover
12.5k21232
edited Aug 15 at 4:15
Math Lover
12.5k21232
12.5k21232
asked Aug 15 at 4:12
H.S.
6
asked Aug 15 at 4:12
H.S.
6
asked Aug 15 at 4:12
H.S.
6
6
5
The result just isn't true. Maybe you wanted even $f$?
– Ian
Aug 15 at 4:19
1
$f(x) = x$ is odd, but $f'(x) = 1$.
– copper.hat
Aug 15 at 5:35
add a comment |Â
5
The result just isn't true. Maybe you wanted even $f$?
– Ian
Aug 15 at 4:19
1
$f(x) = x$ is odd, but $f'(x) = 1$.
– copper.hat
Aug 15 at 5:35
5
5
The result just isn't true. Maybe you wanted even $f$?
– Ian
Aug 15 at 4:19
The result just isn't true. Maybe you wanted even $f$?
– Ian
Aug 15 at 4:19
1
1
$f(x) = x$ is odd, but $f'(x) = 1$.
– copper.hat
Aug 15 at 5:35
$f(x) = x$ is odd, but $f'(x) = 1$.
– copper.hat
Aug 15 at 5:35
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
down vote
Hint:
Consider $f(x) = sin(x)$ and observe that $f'(x) = cos(x)$.
answered Aug 15 at 4:14
Math Lover
12.5k21232
add a comment |Â
up vote
2
down vote
Your way is nice for even $f$;
$f(x) = f(-x) implies f'(x) = -f'(-x)$. Now plugging in $0$ on both sides, we get $f'(0) = -f'(0)$, and solving for $f'(0)$ gives $f'(0) = 0$.
answered Aug 15 at 4:26
Ovi
11.3k935105
add a comment |Â
up vote
1
down vote
As Math Lover points out this isn't true in general; it is if $f(h) = o(h)$ as $hto 0$ though, since in that case
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)+f(h)2h = lim_hto 0fracf(h)h = 0.
$$
This holds for $f(x) = x^2n+1$ for $n>1$, for example. On the other hand if you meant to ask if this holds for even $f$,
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)-f(h)2h = lim_hto 0frac0h = 0
$$
as desired.
answered Aug 15 at 4:23
cdipaolo
677311
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
Hint:
Consider $f(x) = sin(x)$ and observe that $f'(x) = cos(x)$.
answered Aug 15 at 4:14
Math Lover
12.5k21232
add a comment |Â
up vote
7
down vote
Hint:
Consider $f(x) = sin(x)$ and observe that $f'(x) = cos(x)$.
answered Aug 15 at 4:14
Math Lover
12.5k21232
add a comment |Â
up vote
7
down vote
up vote
7
down vote
Hint:
Consider $f(x) = sin(x)$ and observe that $f'(x) = cos(x)$.
answered Aug 15 at 4:14
Math Lover
12.5k21232
Hint:
Consider $f(x) = sin(x)$ and observe that $f'(x) = cos(x)$.
answered Aug 15 at 4:14
Math Lover
12.5k21232
answered Aug 15 at 4:14
Math Lover
12.5k21232
answered Aug 15 at 4:14
Math Lover
12.5k21232
answered Aug 15 at 4:14
Math Lover
12.5k21232
12.5k21232
add a comment |Â
add a comment |Â
up vote
2
down vote
Your way is nice for even $f$;
$f(x) = f(-x) implies f'(x) = -f'(-x)$. Now plugging in $0$ on both sides, we get $f'(0) = -f'(0)$, and solving for $f'(0)$ gives $f'(0) = 0$.
answered Aug 15 at 4:26
Ovi
11.3k935105
add a comment |Â
up vote
2
down vote
Your way is nice for even $f$;
$f(x) = f(-x) implies f'(x) = -f'(-x)$. Now plugging in $0$ on both sides, we get $f'(0) = -f'(0)$, and solving for $f'(0)$ gives $f'(0) = 0$.
answered Aug 15 at 4:26
Ovi
11.3k935105
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your way is nice for even $f$;
$f(x) = f(-x) implies f'(x) = -f'(-x)$. Now plugging in $0$ on both sides, we get $f'(0) = -f'(0)$, and solving for $f'(0)$ gives $f'(0) = 0$.
answered Aug 15 at 4:26
Ovi
11.3k935105
Your way is nice for even $f$;
$f(x) = f(-x) implies f'(x) = -f'(-x)$. Now plugging in $0$ on both sides, we get $f'(0) = -f'(0)$, and solving for $f'(0)$ gives $f'(0) = 0$.
answered Aug 15 at 4:26
Ovi
11.3k935105
answered Aug 15 at 4:26
Ovi
11.3k935105
answered Aug 15 at 4:26
Ovi
11.3k935105
answered Aug 15 at 4:26
Ovi
11.3k935105
11.3k935105
add a comment |Â
add a comment |Â
up vote
1
down vote
As Math Lover points out this isn't true in general; it is if $f(h) = o(h)$ as $hto 0$ though, since in that case
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)+f(h)2h = lim_hto 0fracf(h)h = 0.
$$
This holds for $f(x) = x^2n+1$ for $n>1$, for example. On the other hand if you meant to ask if this holds for even $f$,
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)-f(h)2h = lim_hto 0frac0h = 0
$$
as desired.
answered Aug 15 at 4:23
cdipaolo
677311
add a comment |Â
up vote
1
down vote
As Math Lover points out this isn't true in general; it is if $f(h) = o(h)$ as $hto 0$ though, since in that case
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)+f(h)2h = lim_hto 0fracf(h)h = 0.
$$
This holds for $f(x) = x^2n+1$ for $n>1$, for example. On the other hand if you meant to ask if this holds for even $f$,
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)-f(h)2h = lim_hto 0frac0h = 0
$$
as desired.
answered Aug 15 at 4:23
cdipaolo
677311
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As Math Lover points out this isn't true in general; it is if $f(h) = o(h)$ as $hto 0$ though, since in that case
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)+f(h)2h = lim_hto 0fracf(h)h = 0.
$$
This holds for $f(x) = x^2n+1$ for $n>1$, for example. On the other hand if you meant to ask if this holds for even $f$,
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)-f(h)2h = lim_hto 0frac0h = 0
$$
as desired.
answered Aug 15 at 4:23
cdipaolo
677311
As Math Lover points out this isn't true in general; it is if $f(h) = o(h)$ as $hto 0$ though, since in that case
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)+f(h)2h = lim_hto 0fracf(h)h = 0.
$$
This holds for $f(x) = x^2n+1$ for $n>1$, for example. On the other hand if you meant to ask if this holds for even $f$,
$$
f'(0) = lim_hto 0fracf(h) - f(-h)2h = lim_hto 0fracf(h)-f(h)2h = lim_hto 0frac0h = 0
$$
as desired.
answered Aug 15 at 4:23
cdipaolo
677311
answered Aug 15 at 4:23
cdipaolo
677311
answered Aug 15 at 4:23
cdipaolo
677311
answered Aug 15 at 4:23
cdipaolo
677311
677311
add a comment |Â
add a comment |Â
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5
The result just isn't true. Maybe you wanted even $f$?
– Ian
Aug 15 at 4:19
1
$f(x) = x$ is odd, but $f'(x) = 1$.
– copper.hat
Aug 15 at 5:35