Application of mean value and Rolle's theorems - twice differentiable functions
Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.
I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?
Any help will be appreciated!
calculus rolles-theorem
edited Nov 29 at 18:58
Bernard
118k638112
asked Nov 29 at 18:00
Aishwarya Deore
324
add a comment |
Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.
I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?
Any help will be appreciated!
calculus rolles-theorem
edited Nov 29 at 18:58
Bernard
118k638112
asked Nov 29 at 18:00
Aishwarya Deore
324
add a comment |
Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.
I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?
Any help will be appreciated!
calculus rolles-theorem
edited Nov 29 at 18:58
Bernard
118k638112
asked Nov 29 at 18:00
Aishwarya Deore
324
Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.
I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?
Any help will be appreciated!
calculus rolles-theorem
calculus rolles-theorem
edited Nov 29 at 18:58
Bernard
118k638112
asked Nov 29 at 18:00
Aishwarya Deore
324
edited Nov 29 at 18:58
Bernard
118k638112
asked Nov 29 at 18:00
Aishwarya Deore
324
edited Nov 29 at 18:58
Bernard
118k638112
edited Nov 29 at 18:58
Bernard
118k638112
edited Nov 29 at 18:58
Bernard
118k638112
118k638112
asked Nov 29 at 18:00
Aishwarya Deore
324
asked Nov 29 at 18:00
Aishwarya Deore
324
asked Nov 29 at 18:00
Aishwarya Deore
324
324
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$
By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$
$f''(x)leq0, forall xin(a,b)$
$impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$
$impliesforall xin(a,c), f'(x)geq0$
$implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$
Similarly, $forall xin(c,b), f'(x)leq0$
$impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$
$implies f(x)geq0$
As far as $f(c)$ is concerned, continuity requires $f(c)geq0$
answered Nov 29 at 18:27
Shubham Johri
3,859716
add a comment |
Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
repeatedly:
There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.
There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.
There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.
This is a contradiction to the assumption that $f''(x) le 0$
for all $x in (a, b)$.
An alternative approach: Fix any $c in (a, b)$ and define
$$
g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
$$
Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
Then
$$
0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
$$
Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
a contradiction:
$$
f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
$$
for some $xi in (c, b)$.
The real idea here is that $f$ is concave,
and a continuous concave function on a compact interval attains its
minimum at one of the boundary points: Therefore
$$
f(x) ge min (f(a), f(b)) = 0
$$
for $x in (a, b)$.
answered Nov 29 at 18:17
Martin R
26.9k33251
Thank you! I get the first method
– Aishwarya Deore
Nov 29 at 18:31
add a comment |
By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.
Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so
- if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;
- if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.
Thus, in each case, $f(x)ge 0$.
answered Nov 29 at 19:13
Bernard
118k638112
add a comment |
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3 Answers
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3 Answers
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$f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$
By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$
$f''(x)leq0, forall xin(a,b)$
$impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$
$impliesforall xin(a,c), f'(x)geq0$
$implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$
Similarly, $forall xin(c,b), f'(x)leq0$
$impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$
$implies f(x)geq0$
As far as $f(c)$ is concerned, continuity requires $f(c)geq0$
answered Nov 29 at 18:27
Shubham Johri
3,859716
add a comment |
$f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$
By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$
$f''(x)leq0, forall xin(a,b)$
$impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$
$impliesforall xin(a,c), f'(x)geq0$
$implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$
Similarly, $forall xin(c,b), f'(x)leq0$
$impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$
$implies f(x)geq0$
As far as $f(c)$ is concerned, continuity requires $f(c)geq0$
answered Nov 29 at 18:27
Shubham Johri
3,859716
add a comment |
$f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$
By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$
$f''(x)leq0, forall xin(a,b)$
$impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$
$impliesforall xin(a,c), f'(x)geq0$
$implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$
Similarly, $forall xin(c,b), f'(x)leq0$
$impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$
$implies f(x)geq0$
As far as $f(c)$ is concerned, continuity requires $f(c)geq0$
answered Nov 29 at 18:27
Shubham Johri
3,859716
$f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$
By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$
$f''(x)leq0, forall xin(a,b)$
$impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$
$impliesforall xin(a,c), f'(x)geq0$
$implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$
Similarly, $forall xin(c,b), f'(x)leq0$
$impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$
$implies f(x)geq0$
As far as $f(c)$ is concerned, continuity requires $f(c)geq0$
answered Nov 29 at 18:27
Shubham Johri
3,859716
answered Nov 29 at 18:27
Shubham Johri
3,859716
answered Nov 29 at 18:27
Shubham Johri
3,859716
answered Nov 29 at 18:27
Shubham Johri
3,859716
3,859716
add a comment |
add a comment |
Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
repeatedly:
There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.
There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.
There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.
This is a contradiction to the assumption that $f''(x) le 0$
for all $x in (a, b)$.
An alternative approach: Fix any $c in (a, b)$ and define
$$
g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
$$
Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
Then
$$
0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
$$
Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
a contradiction:
$$
f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
$$
for some $xi in (c, b)$.
The real idea here is that $f$ is concave,
and a continuous concave function on a compact interval attains its
minimum at one of the boundary points: Therefore
$$
f(x) ge min (f(a), f(b)) = 0
$$
for $x in (a, b)$.
answered Nov 29 at 18:17
Martin R
26.9k33251
Thank you! I get the first method
– Aishwarya Deore
Nov 29 at 18:31
add a comment |
Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
repeatedly:
There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.
There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.
There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.
This is a contradiction to the assumption that $f''(x) le 0$
for all $x in (a, b)$.
An alternative approach: Fix any $c in (a, b)$ and define
$$
g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
$$
Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
Then
$$
0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
$$
Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
a contradiction:
$$
f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
$$
for some $xi in (c, b)$.
The real idea here is that $f$ is concave,
and a continuous concave function on a compact interval attains its
minimum at one of the boundary points: Therefore
$$
f(x) ge min (f(a), f(b)) = 0
$$
for $x in (a, b)$.
answered Nov 29 at 18:17
Martin R
26.9k33251
Thank you! I get the first method
– Aishwarya Deore
Nov 29 at 18:31
add a comment |
Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
repeatedly:
There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.
There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.
There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.
This is a contradiction to the assumption that $f''(x) le 0$
for all $x in (a, b)$.
An alternative approach: Fix any $c in (a, b)$ and define
$$
g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
$$
Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
Then
$$
0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
$$
Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
a contradiction:
$$
f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
$$
for some $xi in (c, b)$.
The real idea here is that $f$ is concave,
and a continuous concave function on a compact interval attains its
minimum at one of the boundary points: Therefore
$$
f(x) ge min (f(a), f(b)) = 0
$$
for $x in (a, b)$.
answered Nov 29 at 18:17
Martin R
26.9k33251
Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
repeatedly:
There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.
There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.
There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.
This is a contradiction to the assumption that $f''(x) le 0$
for all $x in (a, b)$.
An alternative approach: Fix any $c in (a, b)$ and define
$$
g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
$$
Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
Then
$$
0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
$$
Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
a contradiction:
$$
f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
$$
for some $xi in (c, b)$.
The real idea here is that $f$ is concave,
and a continuous concave function on a compact interval attains its
minimum at one of the boundary points: Therefore
$$
f(x) ge min (f(a), f(b)) = 0
$$
for $x in (a, b)$.
answered Nov 29 at 18:17
Martin R
26.9k33251
edited Nov 29 at 18:47
answered Nov 29 at 18:17
Martin R
26.9k33251
answered Nov 29 at 18:17
Martin R
26.9k33251
answered Nov 29 at 18:17
Martin R
26.9k33251
26.9k33251
Thank you! I get the first method
– Aishwarya Deore
Nov 29 at 18:31
add a comment |
Thank you! I get the first method
– Aishwarya Deore
Nov 29 at 18:31
Thank you! I get the first method
– Aishwarya Deore
Nov 29 at 18:31
Thank you! I get the first method
– Aishwarya Deore
Nov 29 at 18:31
add a comment |
By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.
Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so
- if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;
- if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.
Thus, in each case, $f(x)ge 0$.
answered Nov 29 at 19:13
Bernard
118k638112
add a comment |
By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.
Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so
- if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;
- if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.
Thus, in each case, $f(x)ge 0$.
answered Nov 29 at 19:13
Bernard
118k638112
add a comment |
By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.
Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so
- if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;
- if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.
Thus, in each case, $f(x)ge 0$.
answered Nov 29 at 19:13
Bernard
118k638112
By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.
Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so
- if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;
- if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.
Thus, in each case, $f(x)ge 0$.
answered Nov 29 at 19:13
Bernard
118k638112
answered Nov 29 at 19:13
Bernard
118k638112
answered Nov 29 at 19:13
Bernard
118k638112
answered Nov 29 at 19:13
Bernard
118k638112
118k638112
add a comment |
add a comment |
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