Is inverse use of mean value theorem right?
$begingroup$
If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then
for any $xin (a,b)$, exists $y, z in [a,b]$, such that
$f '(x)=dfrac{f(z)-f(y)}{z-y}$
Is this right?
analysis derivatives
edited Nov 29 '12 at 16:39
Namaste
1
asked Nov 29 '12 at 16:36
user46262user46262
1144
$endgroup$
add a comment |
$begingroup$
If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then
for any $xin (a,b)$, exists $y, z in [a,b]$, such that
$f '(x)=dfrac{f(z)-f(y)}{z-y}$
Is this right?
analysis derivatives
edited Nov 29 '12 at 16:39
Namaste
1
asked Nov 29 '12 at 16:36
user46262user46262
1144
$endgroup$
add a comment |
$begingroup$
If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then
for any $xin (a,b)$, exists $y, z in [a,b]$, such that
$f '(x)=dfrac{f(z)-f(y)}{z-y}$
Is this right?
analysis derivatives
edited Nov 29 '12 at 16:39
Namaste
1
asked Nov 29 '12 at 16:36
user46262user46262
1144
$endgroup$
If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then
for any $xin (a,b)$, exists $y, z in [a,b]$, such that
$f '(x)=dfrac{f(z)-f(y)}{z-y}$
Is this right?
analysis derivatives
analysis derivatives
edited Nov 29 '12 at 16:39
Namaste
1
asked Nov 29 '12 at 16:36
user46262user46262
1144
edited Nov 29 '12 at 16:39
Namaste
1
asked Nov 29 '12 at 16:36
user46262user46262
1144
edited Nov 29 '12 at 16:39
Namaste
1
edited Nov 29 '12 at 16:39
Namaste
1
edited Nov 29 '12 at 16:39
Namaste
1
1
asked Nov 29 '12 at 16:36
user46262user46262
1144
asked Nov 29 '12 at 16:36
user46262user46262
1144
asked Nov 29 '12 at 16:36
user46262user46262
1144
1144
add a comment |
add a comment |
1 Answer
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$begingroup$
No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
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$begingroup$
No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
$endgroup$
add a comment |
$begingroup$
No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
$endgroup$
add a comment |
$begingroup$
No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
$endgroup$
No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
answered Nov 29 '12 at 16:39
David MitraDavid Mitra
63.7k6102165
63.7k6102165
add a comment |
add a comment |
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