Show that the function $f$ is strictly bounded by an arbitrary number $A$
$begingroup$
Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
$$f(a)<A$$
$$f(x)+f'(x)<A$$
Show that $f(x)<A$ for all $x in (a,b)$.
My attempt: Generate a function $h(x)$ such that
$$h(x)=e^xf(x)$$
$$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
$$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
$$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
$$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.
real-analysis derivatives
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
$$f(a)<A$$
$$f(x)+f'(x)<A$$
Show that $f(x)<A$ for all $x in (a,b)$.
My attempt: Generate a function $h(x)$ such that
$$h(x)=e^xf(x)$$
$$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
$$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
$$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
$$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.
real-analysis derivatives
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
$$f(a)<A$$
$$f(x)+f'(x)<A$$
Show that $f(x)<A$ for all $x in (a,b)$.
My attempt: Generate a function $h(x)$ such that
$$h(x)=e^xf(x)$$
$$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
$$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
$$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
$$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.
real-analysis derivatives
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
$endgroup$
Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
$$f(a)<A$$
$$f(x)+f'(x)<A$$
Show that $f(x)<A$ for all $x in (a,b)$.
My attempt: Generate a function $h(x)$ such that
$$h(x)=e^xf(x)$$
$$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
$$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
$$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
$$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.
real-analysis derivatives
real-analysis derivatives
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
edited Dec 20 '18 at 9:00
weilam06
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
asked Dec 20 '18 at 7:11
weilam06weilam06
30412
30412
add a comment |
add a comment |
1 Answer
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$begingroup$
I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
Then
$$h(c)-h(a)=int_a^c h'(t),dt
<Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
Then
$$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
Then $f(c)<A$.
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
my only worry here is given our definitions in the post $h$ might be defined at $a$?
$endgroup$
– user136920
Dec 20 '18 at 7:42
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
Then
$$h(c)-h(a)=int_a^c h'(t),dt
<Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
Then
$$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
Then $f(c)<A$.
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
my only worry here is given our definitions in the post $h$ might be defined at $a$?
$endgroup$
– user136920
Dec 20 '18 at 7:42
add a comment |
$begingroup$
I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
Then
$$h(c)-h(a)=int_a^c h'(t),dt
<Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
Then
$$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
Then $f(c)<A$.
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
$begingroup$
my only worry here is given our definitions in the post $h$ might be defined at $a$?
$endgroup$
– user136920
Dec 20 '18 at 7:42
add a comment |
$begingroup$
I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
Then
$$h(c)-h(a)=int_a^c h'(t),dt
<Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
Then
$$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
Then $f(c)<A$.
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
$endgroup$
I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
Then
$$h(c)-h(a)=int_a^c h'(t),dt
<Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
Then
$$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
Then $f(c)<A$.
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
answered Dec 20 '18 at 7:21
Lord Shark the UnknownLord Shark the Unknown
105k1160133
105k1160133
$begingroup$
my only worry here is given our definitions in the post $h$ might be defined at $a$?
$endgroup$
– user136920
Dec 20 '18 at 7:42
add a comment |
$begingroup$
my only worry here is given our definitions in the post $h$ might be defined at $a$?
$endgroup$
– user136920
Dec 20 '18 at 7:42
$begingroup$
my only worry here is given our definitions in the post $h$ might be defined at $a$?
$endgroup$
– user136920
Dec 20 '18 at 7:42
$begingroup$
my only worry here is given our definitions in the post $h$ might be defined at $a$?
$endgroup$
– user136920
Dec 20 '18 at 7:42
add a comment |
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