Rolle's theorem and the closed interval
$begingroup$
I think I was/am missing a key concept of Rolle's theorem.
The question says find the value(s) of c that satisfies Rolle's theorem for
$y=-x^2+2$; $[-2,2]$
I see that this is simply a parabola with vertex at $(0,2)$ and that at that point there is possible a horizontal line (slope$=0$).
The theorem states that if $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ and $f(a)=0 $ and $f(b)=0$, then there is at least one point c in the interval $(a,b)$ such that $f'(c)=0$.
So, I first tested to see if $f(-2)=0$ and it does NOT. And further $f(2)$ does NOT equal $0$, so I assumed that I could go no further.
Is that correct? I think maybe what I should do is find the roots of $f(x)$ and if those roots are within $[-2,2]$ [and they ARE], THEN find c.
A little confused on this key concept of Rolle's. Maybe it was a misprint and they meant to put the interval as $[-sqrt (2), sqrt (2)]$ ?
Do the endpoints have to be used in $f(a)$????
calculus
asked Dec 20 '18 at 2:03
user163862user163862
86521016
$endgroup$
add a comment |
$begingroup$
I think I was/am missing a key concept of Rolle's theorem.
The question says find the value(s) of c that satisfies Rolle's theorem for
$y=-x^2+2$; $[-2,2]$
I see that this is simply a parabola with vertex at $(0,2)$ and that at that point there is possible a horizontal line (slope$=0$).
The theorem states that if $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ and $f(a)=0 $ and $f(b)=0$, then there is at least one point c in the interval $(a,b)$ such that $f'(c)=0$.
So, I first tested to see if $f(-2)=0$ and it does NOT. And further $f(2)$ does NOT equal $0$, so I assumed that I could go no further.
Is that correct? I think maybe what I should do is find the roots of $f(x)$ and if those roots are within $[-2,2]$ [and they ARE], THEN find c.
A little confused on this key concept of Rolle's. Maybe it was a misprint and they meant to put the interval as $[-sqrt (2), sqrt (2)]$ ?
Do the endpoints have to be used in $f(a)$????
calculus
asked Dec 20 '18 at 2:03
user163862user163862
86521016
$endgroup$
1
$begingroup$
People often use Rolle's theorem to refer to the case where both endpoints take the same value, regardless of whether that value is 0 or not. You can add / subtract a constant function to see that the two formulations are equivalent.
$endgroup$
– Spencer
Dec 20 '18 at 2:06
$begingroup$
The question is badly written.
$endgroup$
– William Elliot
Dec 20 '18 at 2:12
add a comment |
$begingroup$
I think I was/am missing a key concept of Rolle's theorem.
The question says find the value(s) of c that satisfies Rolle's theorem for
$y=-x^2+2$; $[-2,2]$
I see that this is simply a parabola with vertex at $(0,2)$ and that at that point there is possible a horizontal line (slope$=0$).
The theorem states that if $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ and $f(a)=0 $ and $f(b)=0$, then there is at least one point c in the interval $(a,b)$ such that $f'(c)=0$.
So, I first tested to see if $f(-2)=0$ and it does NOT. And further $f(2)$ does NOT equal $0$, so I assumed that I could go no further.
Is that correct? I think maybe what I should do is find the roots of $f(x)$ and if those roots are within $[-2,2]$ [and they ARE], THEN find c.
A little confused on this key concept of Rolle's. Maybe it was a misprint and they meant to put the interval as $[-sqrt (2), sqrt (2)]$ ?
Do the endpoints have to be used in $f(a)$????
calculus
asked Dec 20 '18 at 2:03
user163862user163862
86521016
$endgroup$
I think I was/am missing a key concept of Rolle's theorem.
The question says find the value(s) of c that satisfies Rolle's theorem for
$y=-x^2+2$; $[-2,2]$
I see that this is simply a parabola with vertex at $(0,2)$ and that at that point there is possible a horizontal line (slope$=0$).
The theorem states that if $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ and $f(a)=0 $ and $f(b)=0$, then there is at least one point c in the interval $(a,b)$ such that $f'(c)=0$.
So, I first tested to see if $f(-2)=0$ and it does NOT. And further $f(2)$ does NOT equal $0$, so I assumed that I could go no further.
Is that correct? I think maybe what I should do is find the roots of $f(x)$ and if those roots are within $[-2,2]$ [and they ARE], THEN find c.
A little confused on this key concept of Rolle's. Maybe it was a misprint and they meant to put the interval as $[-sqrt (2), sqrt (2)]$ ?
Do the endpoints have to be used in $f(a)$????
calculus
calculus
asked Dec 20 '18 at 2:03
user163862user163862
86521016
asked Dec 20 '18 at 2:03
user163862user163862
86521016
asked Dec 20 '18 at 2:03
user163862user163862
86521016
asked Dec 20 '18 at 2:03
user163862user163862
86521016
asked Dec 20 '18 at 2:03
user163862user163862
86521016
86521016
1
$begingroup$
People often use Rolle's theorem to refer to the case where both endpoints take the same value, regardless of whether that value is 0 or not. You can add / subtract a constant function to see that the two formulations are equivalent.
$endgroup$
– Spencer
Dec 20 '18 at 2:06
$begingroup$
The question is badly written.
$endgroup$
– William Elliot
Dec 20 '18 at 2:12
add a comment |
1
$begingroup$
People often use Rolle's theorem to refer to the case where both endpoints take the same value, regardless of whether that value is 0 or not. You can add / subtract a constant function to see that the two formulations are equivalent.
$endgroup$
– Spencer
Dec 20 '18 at 2:06
$begingroup$
The question is badly written.
$endgroup$
– William Elliot
Dec 20 '18 at 2:12
1
1
$begingroup$
People often use Rolle's theorem to refer to the case where both endpoints take the same value, regardless of whether that value is 0 or not. You can add / subtract a constant function to see that the two formulations are equivalent.
$endgroup$
– Spencer
Dec 20 '18 at 2:06
$begingroup$
People often use Rolle's theorem to refer to the case where both endpoints take the same value, regardless of whether that value is 0 or not. You can add / subtract a constant function to see that the two formulations are equivalent.
$endgroup$
– Spencer
Dec 20 '18 at 2:06
$begingroup$
The question is badly written.
$endgroup$
– William Elliot
Dec 20 '18 at 2:12
$begingroup$
The question is badly written.
$endgroup$
– William Elliot
Dec 20 '18 at 2:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is a generalization of Rolle theorem which says that if $f(a)=f(b)$ there exists $c$ such that $f'(c)=0$. Here $f(-2)=f(2)$. This can be easily seen from the previous case, suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$. Define $g$ on $[a,b]$ by $g(x)=f(x)-f(a)$, $g(a)=g(b)=0$ and there exists $c$ such that $g'(c)=f'(c)=0$.
https://en.wikipedia.org/wiki/Rolle%27s_theorem
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
$endgroup$
add a comment |
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1 Answer
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$begingroup$
There is a generalization of Rolle theorem which says that if $f(a)=f(b)$ there exists $c$ such that $f'(c)=0$. Here $f(-2)=f(2)$. This can be easily seen from the previous case, suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$. Define $g$ on $[a,b]$ by $g(x)=f(x)-f(a)$, $g(a)=g(b)=0$ and there exists $c$ such that $g'(c)=f'(c)=0$.
https://en.wikipedia.org/wiki/Rolle%27s_theorem
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
$endgroup$
add a comment |
$begingroup$
There is a generalization of Rolle theorem which says that if $f(a)=f(b)$ there exists $c$ such that $f'(c)=0$. Here $f(-2)=f(2)$. This can be easily seen from the previous case, suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$. Define $g$ on $[a,b]$ by $g(x)=f(x)-f(a)$, $g(a)=g(b)=0$ and there exists $c$ such that $g'(c)=f'(c)=0$.
https://en.wikipedia.org/wiki/Rolle%27s_theorem
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
$endgroup$
add a comment |
$begingroup$
There is a generalization of Rolle theorem which says that if $f(a)=f(b)$ there exists $c$ such that $f'(c)=0$. Here $f(-2)=f(2)$. This can be easily seen from the previous case, suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$. Define $g$ on $[a,b]$ by $g(x)=f(x)-f(a)$, $g(a)=g(b)=0$ and there exists $c$ such that $g'(c)=f'(c)=0$.
https://en.wikipedia.org/wiki/Rolle%27s_theorem
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
$endgroup$
There is a generalization of Rolle theorem which says that if $f(a)=f(b)$ there exists $c$ such that $f'(c)=0$. Here $f(-2)=f(2)$. This can be easily seen from the previous case, suppose that $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and $f(a)=f(b)$. Define $g$ on $[a,b]$ by $g(x)=f(x)-f(a)$, $g(a)=g(b)=0$ and there exists $c$ such that $g'(c)=f'(c)=0$.
https://en.wikipedia.org/wiki/Rolle%27s_theorem
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
answered Dec 20 '18 at 2:08
Tsemo AristideTsemo Aristide
57.4k11444
57.4k11444
add a comment |
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1
$begingroup$
People often use Rolle's theorem to refer to the case where both endpoints take the same value, regardless of whether that value is 0 or not. You can add / subtract a constant function to see that the two formulations are equivalent.
$endgroup$
– Spencer
Dec 20 '18 at 2:06
$begingroup$
The question is badly written.
$endgroup$
– William Elliot
Dec 20 '18 at 2:12