Are derivative functions always bounded on closed intervals?
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Is it true that if a function is differentiable on a closed interval, then its derivative is bounded on that interval?
I know that by Darboux’s theorem, the derivative cannot tend to infinity, but what if it oscillates and the oscillation tends to infinity?
Thanks in advance!
calculus analysis derivatives
asked Jan 9 at 11:54
JiuJiu
552113
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add a comment |
$begingroup$
Is it true that if a function is differentiable on a closed interval, then its derivative is bounded on that interval?
I know that by Darboux’s theorem, the derivative cannot tend to infinity, but what if it oscillates and the oscillation tends to infinity?
Thanks in advance!
calculus analysis derivatives
asked Jan 9 at 11:54
JiuJiu
552113
$endgroup$
add a comment |
$begingroup$
Is it true that if a function is differentiable on a closed interval, then its derivative is bounded on that interval?
I know that by Darboux’s theorem, the derivative cannot tend to infinity, but what if it oscillates and the oscillation tends to infinity?
Thanks in advance!
calculus analysis derivatives
asked Jan 9 at 11:54
JiuJiu
552113
$endgroup$
Is it true that if a function is differentiable on a closed interval, then its derivative is bounded on that interval?
I know that by Darboux’s theorem, the derivative cannot tend to infinity, but what if it oscillates and the oscillation tends to infinity?
Thanks in advance!
calculus analysis derivatives
calculus analysis derivatives
asked Jan 9 at 11:54
JiuJiu
552113
asked Jan 9 at 11:54
JiuJiu
552113
asked Jan 9 at 11:54
JiuJiu
552113
asked Jan 9 at 11:54
JiuJiu
552113
asked Jan 9 at 11:54
JiuJiu
552113
552113
add a comment |
add a comment |
1 Answer
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$f(x)=x^{1.5}sin(frac 1 x)$ for $x neq 0$, $f(0)=0$ is a counter example. Note that $|f'(frac 1 {npi})| to infty$
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
$f(x)=x^{1.5}sin(frac 1 x)$ for $x neq 0$, $f(0)=0$ is a counter example. Note that $|f'(frac 1 {npi})| to infty$
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
$endgroup$
add a comment |
$begingroup$
$f(x)=x^{1.5}sin(frac 1 x)$ for $x neq 0$, $f(0)=0$ is a counter example. Note that $|f'(frac 1 {npi})| to infty$
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
$endgroup$
add a comment |
$begingroup$
$f(x)=x^{1.5}sin(frac 1 x)$ for $x neq 0$, $f(0)=0$ is a counter example. Note that $|f'(frac 1 {npi})| to infty$
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
$endgroup$
$f(x)=x^{1.5}sin(frac 1 x)$ for $x neq 0$, $f(0)=0$ is a counter example. Note that $|f'(frac 1 {npi})| to infty$
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
answered Jan 9 at 11:59
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
75.2k53270
add a comment |
add a comment |
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