Find every differentiable function $f:(0,infty)rightarrow mathbbR$ with the properties $(f(x))^2f'(x)geq x^2$ and $|f(x)|leq x(1+e^-x)$For a differentiable function $f:mathbbR rightarrow mathbbR$ what does $lim_xrightarrow +infty f'(x)=1$ imply? (TIFR GS $2014$)if $f'(x)rightarrow L$ as $ x rightarrow infty$, $-infty leq L leq infty $ then $ f(x)/x rightarrow L $ as $x rightarrow infty$If every composition of a differentiable path and a function is differentiable at 0, means the function is differentiable at 0Function that is continuous and monotone increasing but not differentiable at 0A function who's image is $mathbbR$ on every intervalFind the value of constants $c_1, c_2, c_3, c_4$ for which function $f: mathbbR rightarrow mathbbR$ is differentiableProve that a differentiable function f with $f'(x) geq f(x)^2$ is non positive and has limit $0$Find a $f$ function such that$f'(x)geq 0$ but not continuousExample of function which is twice differentiable with $f,f''$ strictly increasing but $lim_xto inftyf(x)neq infty$Can a continuous real valued function, differentiable everywhere but $x_0$, be expressed as $g(x)+h(x)|x-x_0|$ for some differentiable $g$ and $h$?
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Find every differentiable function $f:(0,infty)rightarrow mathbbR$ with the properties $(f(x))^2f'(x)geq x^2$ and $|f(x)|leq x(1+e^-x)$
For a differentiable function $f:mathbbR rightarrow mathbbR$ what does $lim_xrightarrow +infty f'(x)=1$ imply? (TIFR GS $2014$)if $f'(x)rightarrow L$ as $ x rightarrow infty$, $-infty leq L leq infty $ then $ f(x)/x rightarrow L $ as $x rightarrow infty$If every composition of a differentiable path and a function is differentiable at 0, means the function is differentiable at 0Function that is continuous and monotone increasing but not differentiable at 0A function who's image is $mathbbR$ on every intervalFind the value of constants $c_1, c_2, c_3, c_4$ for which function $f: mathbbR rightarrow mathbbR$ is differentiableProve that a differentiable function f with $f'(x) geq f(x)^2$ is non positive and has limit $0$Find a $f$ function such that$f'(x)geq 0$ but not continuousExample of function which is twice differentiable with $f,f''$ strictly increasing but $lim_xto inftyf(x)neq infty$Can a continuous real valued function, differentiable everywhere but $x_0$, be expressed as $g(x)+h(x)|x-x_0|$ for some differentiable $g$ and $h$?
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I need to find every differentiable function $f:(0,infty)rightarrow mathbbR$ with the properties $(f(x))^2f'(x)geq x^2$ and $|f(x)|leq x(1+e^-x)$, for $xin(0,infty)$. The first property is equivalent to $(f^3(x))'geq 3x^2$, which shows that $f$ is increasing on $(0,infty)$. I tried thinking of a polynomial function, but that was not sufficient. Maybe there is no such function. Any help?
calculus integration derivatives
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add a comment |
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I need to find every differentiable function $f:(0,infty)rightarrow mathbbR$ with the properties $(f(x))^2f'(x)geq x^2$ and $|f(x)|leq x(1+e^-x)$, for $xin(0,infty)$. The first property is equivalent to $(f^3(x))'geq 3x^2$, which shows that $f$ is increasing on $(0,infty)$. I tried thinking of a polynomial function, but that was not sufficient. Maybe there is no such function. Any help?
calculus integration derivatives
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Correction: $f(x)geq f(a) + x- a, forall aleq x$.
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– maxmilgram
Mar 15 at 6:23
add a comment |
$begingroup$
I need to find every differentiable function $f:(0,infty)rightarrow mathbbR$ with the properties $(f(x))^2f'(x)geq x^2$ and $|f(x)|leq x(1+e^-x)$, for $xin(0,infty)$. The first property is equivalent to $(f^3(x))'geq 3x^2$, which shows that $f$ is increasing on $(0,infty)$. I tried thinking of a polynomial function, but that was not sufficient. Maybe there is no such function. Any help?
calculus integration derivatives
$endgroup$
I need to find every differentiable function $f:(0,infty)rightarrow mathbbR$ with the properties $(f(x))^2f'(x)geq x^2$ and $|f(x)|leq x(1+e^-x)$, for $xin(0,infty)$. The first property is equivalent to $(f^3(x))'geq 3x^2$, which shows that $f$ is increasing on $(0,infty)$. I tried thinking of a polynomial function, but that was not sufficient. Maybe there is no such function. Any help?
calculus integration derivatives
calculus integration derivatives
edited Mar 15 at 6:21
asked Mar 15 at 6:11
user651692
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Correction: $f(x)geq f(a) + x- a, forall aleq x$.
$endgroup$
– maxmilgram
Mar 15 at 6:23
add a comment |
$begingroup$
Correction: $f(x)geq f(a) + x- a, forall aleq x$.
$endgroup$
– maxmilgram
Mar 15 at 6:23
$begingroup$
Correction: $f(x)geq f(a) + x- a, forall aleq x$.
$endgroup$
– maxmilgram
Mar 15 at 6:23
$begingroup$
Correction: $f(x)geq f(a) + x- a, forall aleq x$.
$endgroup$
– maxmilgram
Mar 15 at 6:23
add a comment |
1 Answer
1
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Define $g(x)=f(x)^3-x^3$. Observe that $g$ is increasing since $g'(x)=3f(x)^2f'(x)-3x^2ge 0$, and $displaystyle lim_xto 0^+ g(x) = lim_xto 0^+ f(x) = 0$ by the given condition $|f(x)|le x(1+e^-x)$. Since $g$ is increasing, for each $x$, we have
$$
0le g(x)le lim_ytoinftyg(y)= lim_ytoinftyf(y)^3-y^3 le lim_ytoinfty y^3(3e^-y+3e^-2y+e^-3y)=0,
$$ which implies $g(x)=f(x)^3-x^3=0$. Thus we obtain $f(x)=x$.
answered Mar 15 at 6:22
SongSong
18.5k21651
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$begingroup$
Define $g(x)=f(x)^3-x^3$. Observe that $g$ is increasing since $g'(x)=3f(x)^2f'(x)-3x^2ge 0$, and $displaystyle lim_xto 0^+ g(x) = lim_xto 0^+ f(x) = 0$ by the given condition $|f(x)|le x(1+e^-x)$. Since $g$ is increasing, for each $x$, we have
$$
0le g(x)le lim_ytoinftyg(y)= lim_ytoinftyf(y)^3-y^3 le lim_ytoinfty y^3(3e^-y+3e^-2y+e^-3y)=0,
$$ which implies $g(x)=f(x)^3-x^3=0$. Thus we obtain $f(x)=x$.
answered Mar 15 at 6:22
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Define $g(x)=f(x)^3-x^3$. Observe that $g$ is increasing since $g'(x)=3f(x)^2f'(x)-3x^2ge 0$, and $displaystyle lim_xto 0^+ g(x) = lim_xto 0^+ f(x) = 0$ by the given condition $|f(x)|le x(1+e^-x)$. Since $g$ is increasing, for each $x$, we have
$$
0le g(x)le lim_ytoinftyg(y)= lim_ytoinftyf(y)^3-y^3 le lim_ytoinfty y^3(3e^-y+3e^-2y+e^-3y)=0,
$$ which implies $g(x)=f(x)^3-x^3=0$. Thus we obtain $f(x)=x$.
answered Mar 15 at 6:22
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
Define $g(x)=f(x)^3-x^3$. Observe that $g$ is increasing since $g'(x)=3f(x)^2f'(x)-3x^2ge 0$, and $displaystyle lim_xto 0^+ g(x) = lim_xto 0^+ f(x) = 0$ by the given condition $|f(x)|le x(1+e^-x)$. Since $g$ is increasing, for each $x$, we have
$$
0le g(x)le lim_ytoinftyg(y)= lim_ytoinftyf(y)^3-y^3 le lim_ytoinfty y^3(3e^-y+3e^-2y+e^-3y)=0,
$$ which implies $g(x)=f(x)^3-x^3=0$. Thus we obtain $f(x)=x$.
answered Mar 15 at 6:22
SongSong
18.5k21651
$endgroup$
Define $g(x)=f(x)^3-x^3$. Observe that $g$ is increasing since $g'(x)=3f(x)^2f'(x)-3x^2ge 0$, and $displaystyle lim_xto 0^+ g(x) = lim_xto 0^+ f(x) = 0$ by the given condition $|f(x)|le x(1+e^-x)$. Since $g$ is increasing, for each $x$, we have
$$
0le g(x)le lim_ytoinftyg(y)= lim_ytoinftyf(y)^3-y^3 le lim_ytoinfty y^3(3e^-y+3e^-2y+e^-3y)=0,
$$ which implies $g(x)=f(x)^3-x^3=0$. Thus we obtain $f(x)=x$.
answered Mar 15 at 6:22
SongSong
18.5k21651
answered Mar 15 at 6:22
SongSong
18.5k21651
answered Mar 15 at 6:22
SongSong
18.5k21651
answered Mar 15 at 6:22
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
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$begingroup$
Correction: $f(x)geq f(a) + x- a, forall aleq x$.
$endgroup$
– maxmilgram
Mar 15 at 6:23