Mixing
Proving a function is less or equal to a polynomial
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Lets say we got a differentiable function F: ℝ → ℝ and f(0) = a, f'(x) ≤ b for any x > 0.
Now, how to prove f(x) ≤ bx + a for any x ≥ 0 ?
It seems to be kinda simple, but I'm stuck.
real-analysis functions derivatives
asked Feb 1 at 17:00
sddssdds
316
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add a comment |
$begingroup$
Lets say we got a differentiable function F: ℝ → ℝ and f(0) = a, f'(x) ≤ b for any x > 0.
Now, how to prove f(x) ≤ bx + a for any x ≥ 0 ?
It seems to be kinda simple, but I'm stuck.
real-analysis functions derivatives
asked Feb 1 at 17:00
sddssdds
316
$endgroup$
add a comment |
$begingroup$
Lets say we got a differentiable function F: ℝ → ℝ and f(0) = a, f'(x) ≤ b for any x > 0.
Now, how to prove f(x) ≤ bx + a for any x ≥ 0 ?
It seems to be kinda simple, but I'm stuck.
real-analysis functions derivatives
asked Feb 1 at 17:00
sddssdds
316
$endgroup$
Lets say we got a differentiable function F: ℝ → ℝ and f(0) = a, f'(x) ≤ b for any x > 0.
Now, how to prove f(x) ≤ bx + a for any x ≥ 0 ?
It seems to be kinda simple, but I'm stuck.
real-analysis functions derivatives
real-analysis functions derivatives
asked Feb 1 at 17:00
sddssdds
316
asked Feb 1 at 17:00
sddssdds
316
asked Feb 1 at 17:00
sddssdds
316
asked Feb 1 at 17:00
sddssdds
316
asked Feb 1 at 17:00
sddssdds
316
316
add a comment |
add a comment |
4 Answers
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Let $g(x) = bx + a$. Consider the function $h(x) = g(x) - f(x)$. We have $h(0) = 0$ and $h'(x) ge 0$ for all $x > 0$. Therefore, $h$ is non-decreasing on $(0,infty)$ and since $h(0) = 0$, we must have $h(x) ge 0$ for all $xge 0$. In other words, $g(x) ge f(x)$ for all $xge 0$.
answered Feb 1 at 17:03
cspruncsprun
2,804211
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Integrals are monotone, hence you can just integrate $f'(x) leq b$.
answered Feb 1 at 17:04
KlausKlaus
2,955214
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Given $xgt 0$, we can find $xiin (0,x)$ such that $$f'(xi)=frac {f (x)-f (0)}{x-0}.$$ Since $xigt 0$, $f'(xi)leq b$,$implies dfrac {f (x)-f (0)}{x-0}leq b $. Substituting $f (0)=a $, we get $f (x)leq bx+a $.
$x=0$ case is trivial.
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
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If $h(x)le g(x)$ then $int_a^b h(x)dx le int_a^b g(x)dx$
answered Feb 1 at 17:06
VasyaVasya
4,3571618
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
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active
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votes
$begingroup$
Let $g(x) = bx + a$. Consider the function $h(x) = g(x) - f(x)$. We have $h(0) = 0$ and $h'(x) ge 0$ for all $x > 0$. Therefore, $h$ is non-decreasing on $(0,infty)$ and since $h(0) = 0$, we must have $h(x) ge 0$ for all $xge 0$. In other words, $g(x) ge f(x)$ for all $xge 0$.
answered Feb 1 at 17:03
cspruncsprun
2,804211
$endgroup$
add a comment |
$begingroup$
Let $g(x) = bx + a$. Consider the function $h(x) = g(x) - f(x)$. We have $h(0) = 0$ and $h'(x) ge 0$ for all $x > 0$. Therefore, $h$ is non-decreasing on $(0,infty)$ and since $h(0) = 0$, we must have $h(x) ge 0$ for all $xge 0$. In other words, $g(x) ge f(x)$ for all $xge 0$.
answered Feb 1 at 17:03
cspruncsprun
2,804211
$endgroup$
add a comment |
$begingroup$
Let $g(x) = bx + a$. Consider the function $h(x) = g(x) - f(x)$. We have $h(0) = 0$ and $h'(x) ge 0$ for all $x > 0$. Therefore, $h$ is non-decreasing on $(0,infty)$ and since $h(0) = 0$, we must have $h(x) ge 0$ for all $xge 0$. In other words, $g(x) ge f(x)$ for all $xge 0$.
answered Feb 1 at 17:03
cspruncsprun
2,804211
$endgroup$
Let $g(x) = bx + a$. Consider the function $h(x) = g(x) - f(x)$. We have $h(0) = 0$ and $h'(x) ge 0$ for all $x > 0$. Therefore, $h$ is non-decreasing on $(0,infty)$ and since $h(0) = 0$, we must have $h(x) ge 0$ for all $xge 0$. In other words, $g(x) ge f(x)$ for all $xge 0$.
answered Feb 1 at 17:03
cspruncsprun
2,804211
answered Feb 1 at 17:03
cspruncsprun
2,804211
answered Feb 1 at 17:03
cspruncsprun
2,804211
answered Feb 1 at 17:03
cspruncsprun
2,804211
2,804211
add a comment |
add a comment |
$begingroup$
Integrals are monotone, hence you can just integrate $f'(x) leq b$.
answered Feb 1 at 17:04
KlausKlaus
2,955214
$endgroup$
add a comment |
$begingroup$
Integrals are monotone, hence you can just integrate $f'(x) leq b$.
answered Feb 1 at 17:04
KlausKlaus
2,955214
$endgroup$
add a comment |
$begingroup$
Integrals are monotone, hence you can just integrate $f'(x) leq b$.
answered Feb 1 at 17:04
KlausKlaus
2,955214
$endgroup$
Integrals are monotone, hence you can just integrate $f'(x) leq b$.
answered Feb 1 at 17:04
KlausKlaus
2,955214
answered Feb 1 at 17:04
KlausKlaus
2,955214
answered Feb 1 at 17:04
KlausKlaus
2,955214
answered Feb 1 at 17:04
KlausKlaus
2,955214
2,955214
add a comment |
add a comment |
$begingroup$
Given $xgt 0$, we can find $xiin (0,x)$ such that $$f'(xi)=frac {f (x)-f (0)}{x-0}.$$ Since $xigt 0$, $f'(xi)leq b$,$implies dfrac {f (x)-f (0)}{x-0}leq b $. Substituting $f (0)=a $, we get $f (x)leq bx+a $.
$x=0$ case is trivial.
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
add a comment |
$begingroup$
Given $xgt 0$, we can find $xiin (0,x)$ such that $$f'(xi)=frac {f (x)-f (0)}{x-0}.$$ Since $xigt 0$, $f'(xi)leq b$,$implies dfrac {f (x)-f (0)}{x-0}leq b $. Substituting $f (0)=a $, we get $f (x)leq bx+a $.
$x=0$ case is trivial.
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
add a comment |
$begingroup$
Given $xgt 0$, we can find $xiin (0,x)$ such that $$f'(xi)=frac {f (x)-f (0)}{x-0}.$$ Since $xigt 0$, $f'(xi)leq b$,$implies dfrac {f (x)-f (0)}{x-0}leq b $. Substituting $f (0)=a $, we get $f (x)leq bx+a $.
$x=0$ case is trivial.
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
$endgroup$
Given $xgt 0$, we can find $xiin (0,x)$ such that $$f'(xi)=frac {f (x)-f (0)}{x-0}.$$ Since $xigt 0$, $f'(xi)leq b$,$implies dfrac {f (x)-f (0)}{x-0}leq b $. Substituting $f (0)=a $, we get $f (x)leq bx+a $.
$x=0$ case is trivial.
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
answered Feb 1 at 17:13
Thomas ShelbyThomas Shelby
4,7362727
4,7362727
add a comment |
add a comment |
$begingroup$
If $h(x)le g(x)$ then $int_a^b h(x)dx le int_a^b g(x)dx$
answered Feb 1 at 17:06
VasyaVasya
4,3571618
$endgroup$
add a comment |
$begingroup$
If $h(x)le g(x)$ then $int_a^b h(x)dx le int_a^b g(x)dx$
answered Feb 1 at 17:06
VasyaVasya
4,3571618
$endgroup$
add a comment |
$begingroup$
If $h(x)le g(x)$ then $int_a^b h(x)dx le int_a^b g(x)dx$
answered Feb 1 at 17:06
VasyaVasya
4,3571618
$endgroup$
If $h(x)le g(x)$ then $int_a^b h(x)dx le int_a^b g(x)dx$
answered Feb 1 at 17:06
VasyaVasya
4,3571618
answered Feb 1 at 17:06
VasyaVasya
4,3571618
answered Feb 1 at 17:06
VasyaVasya
4,3571618
answered Feb 1 at 17:06
VasyaVasya
4,3571618
4,3571618
add a comment |
add a comment |
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