Mixing
How to construct a function with desired properties?
$begingroup$
I was wondering if there is any systematic way to come up with a function with specific properties.
I want to have a concave function in domain of $[0, 1]$ where $f(0) = 0$ and $f(1) = 1$. In other words it is concave down, but it crosses $f(x) = x$ at $x = 0$ and $x = 1$. I also want to have a parameter that controls the concavity.
I was thinking of $-kx^2 + (k+1)x$ with trial and error.
In general, what is the best way to come up with a function having specific properties?
functions graphing-functions
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
asked Jan 24 at 5:41
ShabnamShabnam
134
$endgroup$
add a comment |
$begingroup$
I was wondering if there is any systematic way to come up with a function with specific properties.
I want to have a concave function in domain of $[0, 1]$ where $f(0) = 0$ and $f(1) = 1$. In other words it is concave down, but it crosses $f(x) = x$ at $x = 0$ and $x = 1$. I also want to have a parameter that controls the concavity.
I was thinking of $-kx^2 + (k+1)x$ with trial and error.
In general, what is the best way to come up with a function having specific properties?
functions graphing-functions
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
asked Jan 24 at 5:41
ShabnamShabnam
134
$endgroup$
add a comment |
$begingroup$
I was wondering if there is any systematic way to come up with a function with specific properties.
I want to have a concave function in domain of $[0, 1]$ where $f(0) = 0$ and $f(1) = 1$. In other words it is concave down, but it crosses $f(x) = x$ at $x = 0$ and $x = 1$. I also want to have a parameter that controls the concavity.
I was thinking of $-kx^2 + (k+1)x$ with trial and error.
In general, what is the best way to come up with a function having specific properties?
functions graphing-functions
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
asked Jan 24 at 5:41
ShabnamShabnam
134
$endgroup$
I was wondering if there is any systematic way to come up with a function with specific properties.
I want to have a concave function in domain of $[0, 1]$ where $f(0) = 0$ and $f(1) = 1$. In other words it is concave down, but it crosses $f(x) = x$ at $x = 0$ and $x = 1$. I also want to have a parameter that controls the concavity.
I was thinking of $-kx^2 + (k+1)x$ with trial and error.
In general, what is the best way to come up with a function having specific properties?
functions graphing-functions
functions graphing-functions
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
asked Jan 24 at 5:41
ShabnamShabnam
134
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
asked Jan 24 at 5:41
ShabnamShabnam
134
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
edited Jan 24 at 5:46
Lucas Corrêa
1,5751421
1,5751421
asked Jan 24 at 5:41
ShabnamShabnam
134
asked Jan 24 at 5:41
ShabnamShabnam
134
asked Jan 24 at 5:41
ShabnamShabnam
134
134
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $f:[0,1] to mathbb R$ is twice differentiable, then we have:
$f$ is concave $ iff f'' le 0 $ on $[0,1]$.
Now let $f$ be of the form $f(x)=ax^2+bx+c$. Then it is easy to see, that $f$ is concave, $f(0)=0 $ and $f(1)=1 iff a le 0$ and $b=1-a.$
answered Jan 24 at 5:56
FredFred
48.3k1849
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $f:[0,1] to mathbb R$ is twice differentiable, then we have:
$f$ is concave $ iff f'' le 0 $ on $[0,1]$.
Now let $f$ be of the form $f(x)=ax^2+bx+c$. Then it is easy to see, that $f$ is concave, $f(0)=0 $ and $f(1)=1 iff a le 0$ and $b=1-a.$
answered Jan 24 at 5:56
FredFred
48.3k1849
$endgroup$
add a comment |
$begingroup$
If $f:[0,1] to mathbb R$ is twice differentiable, then we have:
$f$ is concave $ iff f'' le 0 $ on $[0,1]$.
Now let $f$ be of the form $f(x)=ax^2+bx+c$. Then it is easy to see, that $f$ is concave, $f(0)=0 $ and $f(1)=1 iff a le 0$ and $b=1-a.$
answered Jan 24 at 5:56
FredFred
48.3k1849
$endgroup$
add a comment |
$begingroup$
If $f:[0,1] to mathbb R$ is twice differentiable, then we have:
$f$ is concave $ iff f'' le 0 $ on $[0,1]$.
Now let $f$ be of the form $f(x)=ax^2+bx+c$. Then it is easy to see, that $f$ is concave, $f(0)=0 $ and $f(1)=1 iff a le 0$ and $b=1-a.$
answered Jan 24 at 5:56
FredFred
48.3k1849
$endgroup$
If $f:[0,1] to mathbb R$ is twice differentiable, then we have:
$f$ is concave $ iff f'' le 0 $ on $[0,1]$.
Now let $f$ be of the form $f(x)=ax^2+bx+c$. Then it is easy to see, that $f$ is concave, $f(0)=0 $ and $f(1)=1 iff a le 0$ and $b=1-a.$
answered Jan 24 at 5:56
FredFred
48.3k1849
answered Jan 24 at 5:56
FredFred
48.3k1849
answered Jan 24 at 5:56
FredFred
48.3k1849
answered Jan 24 at 5:56
FredFred
48.3k1849
48.3k1849
add a comment |
add a comment |
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