Show $f(x) >0$ for $x>x_0$ if its $f' >f$ and $f(x_0)=0$
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Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.
Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.
calculus
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Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.
Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.
calculus
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You can use strict inequality.
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– user25959
Nov 24 '18 at 16:40
add a comment |
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Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.
Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.
calculus
$endgroup$
Let $f: mathbb{R} rightarrow mathbb{R}$ be a differentiable function. Suppose that $f'(x)>f(x)$ for all $x in mathbb{R}$, and $f(x_0)=0$ for some $x_0 in R$. Prove that $f(x)>0$ for all $x>x_0$. As an application of this, show that $ae^x=a+x+x^2/2$.
Here is my attempt:
We know that $lim_{h to 0}{frac{f(x+h)-f(x)}{h}}-f(x)geq 0$. So let $x=x_o+h$, $lim_{h to 0}{frac{f(x_0+h)-f(x_0)}{h}}-f(x_0)geq 0$. So $lim_{h to 0}{frac{f(x_0+h)}{h}}geq 0$. And I'm stuck.
calculus
calculus
asked Nov 24 '18 at 16:37
nafhgoodnafhgood
1,803422
1,803422
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You can use strict inequality.
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– user25959
Nov 24 '18 at 16:40
add a comment |
$begingroup$
You can use strict inequality.
$endgroup$
– user25959
Nov 24 '18 at 16:40
$begingroup$
You can use strict inequality.
$endgroup$
– user25959
Nov 24 '18 at 16:40
$begingroup$
You can use strict inequality.
$endgroup$
– user25959
Nov 24 '18 at 16:40
add a comment |
2 Answers
2
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oldest
votes
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Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.
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For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
$$
g(x) - g(x_0) = g'(c) (x-x_0)
$$
i.e.
$$
e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
$$
so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.
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Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
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– user25959
Nov 24 '18 at 16:56
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Thank you, I have corrected the answer.
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– Rigel
Nov 24 '18 at 17:07
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.
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add a comment |
$begingroup$
Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.
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add a comment |
$begingroup$
Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.
$endgroup$
Proceed by contradiction. First since $f'(x_0) > f(x_0) = 0$, there is some $t > 0$ s.t. $f (x)>0$ on $(x_0, x_0+t)$. Assume $f(x) leqslant 0$ for some $z > x_0$, and $z$ is the minimal one, i.e. $f(y) >0$ when $x_0 < y < z$, hence $f'(y) >0$ on $(x_0, z)$ as well, and by the MVT, $f(z) > f(x_0) = 0$, contradiction.
answered Nov 24 '18 at 16:48
xbhxbh
5,9781522
5,9781522
add a comment |
add a comment |
$begingroup$
For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
$$
g(x) - g(x_0) = g'(c) (x-x_0)
$$
i.e.
$$
e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
$$
so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.
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Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
$endgroup$
– user25959
Nov 24 '18 at 16:56
$begingroup$
Thank you, I have corrected the answer.
$endgroup$
– Rigel
Nov 24 '18 at 17:07
add a comment |
$begingroup$
For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
$$
g(x) - g(x_0) = g'(c) (x-x_0)
$$
i.e.
$$
e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
$$
so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.
$endgroup$
$begingroup$
Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
$endgroup$
– user25959
Nov 24 '18 at 16:56
$begingroup$
Thank you, I have corrected the answer.
$endgroup$
– Rigel
Nov 24 '18 at 17:07
add a comment |
$begingroup$
For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
$$
g(x) - g(x_0) = g'(c) (x-x_0)
$$
i.e.
$$
e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
$$
so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.
$endgroup$
For the first part: let $g(x) = e^{-x}{f(x)}$ and let $x > x_0$. By the MVT there exists $c in (x_0, x)$ such that
$$
g(x) - g(x_0) = g'(c) (x-x_0)
$$
i.e.
$$
e^{-x}{f(x)} - e^{-x_0}{f(x_0)} = e^{-c}[f'(c) - f(c)] (x-x_0) > 0
$$
so that $e^{-x}f(x) > 0$, i.e. $f(x) > 0$.
edited Nov 24 '18 at 17:16
answered Nov 24 '18 at 16:45
RigelRigel
10.9k11320
10.9k11320
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Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
$endgroup$
– user25959
Nov 24 '18 at 16:56
$begingroup$
Thank you, I have corrected the answer.
$endgroup$
– Rigel
Nov 24 '18 at 17:07
add a comment |
$begingroup$
Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
$endgroup$
– user25959
Nov 24 '18 at 16:56
$begingroup$
Thank you, I have corrected the answer.
$endgroup$
– Rigel
Nov 24 '18 at 17:07
$begingroup$
Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
$endgroup$
– user25959
Nov 24 '18 at 16:56
$begingroup$
Problem doesn't say $f'>0$ so we don't know a priori that $f'(c)>0$
$endgroup$
– user25959
Nov 24 '18 at 16:56
$begingroup$
Thank you, I have corrected the answer.
$endgroup$
– Rigel
Nov 24 '18 at 17:07
$begingroup$
Thank you, I have corrected the answer.
$endgroup$
– Rigel
Nov 24 '18 at 17:07
add a comment |
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$begingroup$
You can use strict inequality.
$endgroup$
– user25959
Nov 24 '18 at 16:40