How to determine $f(0)$?
$$f(ax+b) = x$$
$$f(a) = dfrac{b}{a}$$
I want to determine $f(0)$ from given equations.
My attempt:
$$f(a)= dfrac{b}{a} implies f(0) = dfrac{b}{0} = text{undefined} $$
However, directly determining $f(0)$ makes no sense in this case. Could you assist me with that?
functions
asked Nov 25 '18 at 17:12
Enzo
996
|
show 2 more comments
$$f(ax+b) = x$$
$$f(a) = dfrac{b}{a}$$
I want to determine $f(0)$ from given equations.
My attempt:
$$f(a)= dfrac{b}{a} implies f(0) = dfrac{b}{0} = text{undefined} $$
However, directly determining $f(0)$ makes no sense in this case. Could you assist me with that?
functions
asked Nov 25 '18 at 17:12
Enzo
996
1
Substitute $x=dfrac{-b}{a}$.
– Yadati Kiran
Nov 25 '18 at 17:14
Surely $a$ is a constant here and not a variable?
– Mark Bennet
Nov 25 '18 at 17:16
@Enzo Not possible to find a numberical value from the facts given. Also, if $a$ is a constant then using $a=0$ is wrong. You must determine $f(0)$ from the initial expression.
– Rebellos
Nov 25 '18 at 17:19
@Rebellos Assume that $a, b neq 0 in mathcal{R}$, Is it the same in this case?
– Enzo
Nov 25 '18 at 17:19
@Enzo Yes. Also if you want to typeset the set of real numbers, try mathbb R which produces $mathbb R$.
– Rebellos
Nov 25 '18 at 17:22
|
show 2 more comments
$$f(ax+b) = x$$
$$f(a) = dfrac{b}{a}$$
I want to determine $f(0)$ from given equations.
My attempt:
$$f(a)= dfrac{b}{a} implies f(0) = dfrac{b}{0} = text{undefined} $$
However, directly determining $f(0)$ makes no sense in this case. Could you assist me with that?
functions
asked Nov 25 '18 at 17:12
Enzo
996
$$f(ax+b) = x$$
$$f(a) = dfrac{b}{a}$$
I want to determine $f(0)$ from given equations.
My attempt:
$$f(a)= dfrac{b}{a} implies f(0) = dfrac{b}{0} = text{undefined} $$
However, directly determining $f(0)$ makes no sense in this case. Could you assist me with that?
functions
functions
asked Nov 25 '18 at 17:12
Enzo
996
asked Nov 25 '18 at 17:12
Enzo
996
edited Nov 25 '18 at 17:14
asked Nov 25 '18 at 17:12
Enzo
996
asked Nov 25 '18 at 17:12
Enzo
996
asked Nov 25 '18 at 17:12
Enzo
996
996
1
Substitute $x=dfrac{-b}{a}$.
– Yadati Kiran
Nov 25 '18 at 17:14
Surely $a$ is a constant here and not a variable?
– Mark Bennet
Nov 25 '18 at 17:16
@Enzo Not possible to find a numberical value from the facts given. Also, if $a$ is a constant then using $a=0$ is wrong. You must determine $f(0)$ from the initial expression.
– Rebellos
Nov 25 '18 at 17:19
@Rebellos Assume that $a, b neq 0 in mathcal{R}$, Is it the same in this case?
– Enzo
Nov 25 '18 at 17:19
@Enzo Yes. Also if you want to typeset the set of real numbers, try mathbb R which produces $mathbb R$.
– Rebellos
Nov 25 '18 at 17:22
|
show 2 more comments
1
Substitute $x=dfrac{-b}{a}$.
– Yadati Kiran
Nov 25 '18 at 17:14
Surely $a$ is a constant here and not a variable?
– Mark Bennet
Nov 25 '18 at 17:16
@Enzo Not possible to find a numberical value from the facts given. Also, if $a$ is a constant then using $a=0$ is wrong. You must determine $f(0)$ from the initial expression.
– Rebellos
Nov 25 '18 at 17:19
@Rebellos Assume that $a, b neq 0 in mathcal{R}$, Is it the same in this case?
– Enzo
Nov 25 '18 at 17:19
@Enzo Yes. Also if you want to typeset the set of real numbers, try mathbb R which produces $mathbb R$.
– Rebellos
Nov 25 '18 at 17:22
1
1
Substitute $x=dfrac{-b}{a}$.
– Yadati Kiran
Nov 25 '18 at 17:14
Substitute $x=dfrac{-b}{a}$.
– Yadati Kiran
Nov 25 '18 at 17:14
Surely $a$ is a constant here and not a variable?
– Mark Bennet
Nov 25 '18 at 17:16
Surely $a$ is a constant here and not a variable?
– Mark Bennet
Nov 25 '18 at 17:16
@Enzo Not possible to find a numberical value from the facts given. Also, if $a$ is a constant then using $a=0$ is wrong. You must determine $f(0)$ from the initial expression.
– Rebellos
Nov 25 '18 at 17:19
@Enzo Not possible to find a numberical value from the facts given. Also, if $a$ is a constant then using $a=0$ is wrong. You must determine $f(0)$ from the initial expression.
– Rebellos
Nov 25 '18 at 17:19
@Rebellos Assume that $a, b neq 0 in mathcal{R}$, Is it the same in this case?
– Enzo
Nov 25 '18 at 17:19
@Rebellos Assume that $a, b neq 0 in mathcal{R}$, Is it the same in this case?
– Enzo
Nov 25 '18 at 17:19
@Enzo Yes. Also if you want to typeset the set of real numbers, try mathbb R which produces $mathbb R$.
– Rebellos
Nov 25 '18 at 17:22
@Enzo Yes. Also if you want to typeset the set of real numbers, try mathbb R which produces $mathbb R$.
– Rebellos
Nov 25 '18 at 17:22
|
show 2 more comments
3 Answers
3
active
oldest
votes
By the way upon computing, the function is $f(x)=dfrac{b(2x-a)}{a^2}$ and $f(0)=dfrac{-b}{a}$. Further using $f(ax+b)=x$ we get $b=dfrac{a}{2}$. So the function is $f(x)=dfrac{2x-a}{2a}$ and $f(0)=dfrac{-1}{2}$, $f(a)=dfrac{1}{2}$.
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
add a comment |
Well, you want the argument of your given function to be equal to zero. Thus, considering the argument of the funtion $f(ax+b) = x$, we have :
$$ax + b = 0 Leftrightarrow x = -frac{b}{a}$$
Then, $f(0)$ would be yielded for $x= -b/a$, thus again by substituting in the first function expression :
$$f(0)=-frac{b}{a}=-f(a)$$
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
But is it possible to find a numerical value?
– Enzo
Nov 25 '18 at 17:17
@Enzo No, since all your given expressions involve arbitrary constants.
– Rebellos
Nov 25 '18 at 17:18
add a comment |
Here is a start - and contrary to comments, there is a solution.
Let $y=ax+b$ so that $f(y)=x=cfrac {y-b}a$ so now we have expressed $f$ as a linear function of its argument.
I have not yet used the value for $f(a)$ that has been given, so this is the next step.
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
add a comment |
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3 Answers
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oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
By the way upon computing, the function is $f(x)=dfrac{b(2x-a)}{a^2}$ and $f(0)=dfrac{-b}{a}$. Further using $f(ax+b)=x$ we get $b=dfrac{a}{2}$. So the function is $f(x)=dfrac{2x-a}{2a}$ and $f(0)=dfrac{-1}{2}$, $f(a)=dfrac{1}{2}$.
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
add a comment |
By the way upon computing, the function is $f(x)=dfrac{b(2x-a)}{a^2}$ and $f(0)=dfrac{-b}{a}$. Further using $f(ax+b)=x$ we get $b=dfrac{a}{2}$. So the function is $f(x)=dfrac{2x-a}{2a}$ and $f(0)=dfrac{-1}{2}$, $f(a)=dfrac{1}{2}$.
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
add a comment |
By the way upon computing, the function is $f(x)=dfrac{b(2x-a)}{a^2}$ and $f(0)=dfrac{-b}{a}$. Further using $f(ax+b)=x$ we get $b=dfrac{a}{2}$. So the function is $f(x)=dfrac{2x-a}{2a}$ and $f(0)=dfrac{-1}{2}$, $f(a)=dfrac{1}{2}$.
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
By the way upon computing, the function is $f(x)=dfrac{b(2x-a)}{a^2}$ and $f(0)=dfrac{-b}{a}$. Further using $f(ax+b)=x$ we get $b=dfrac{a}{2}$. So the function is $f(x)=dfrac{2x-a}{2a}$ and $f(0)=dfrac{-1}{2}$, $f(a)=dfrac{1}{2}$.
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
edited Nov 25 '18 at 17:47
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
answered Nov 25 '18 at 17:39
Yadati Kiran
1,691619
1,691619
add a comment |
add a comment |
Well, you want the argument of your given function to be equal to zero. Thus, considering the argument of the funtion $f(ax+b) = x$, we have :
$$ax + b = 0 Leftrightarrow x = -frac{b}{a}$$
Then, $f(0)$ would be yielded for $x= -b/a$, thus again by substituting in the first function expression :
$$f(0)=-frac{b}{a}=-f(a)$$
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
But is it possible to find a numerical value?
– Enzo
Nov 25 '18 at 17:17
@Enzo No, since all your given expressions involve arbitrary constants.
– Rebellos
Nov 25 '18 at 17:18
add a comment |
Well, you want the argument of your given function to be equal to zero. Thus, considering the argument of the funtion $f(ax+b) = x$, we have :
$$ax + b = 0 Leftrightarrow x = -frac{b}{a}$$
Then, $f(0)$ would be yielded for $x= -b/a$, thus again by substituting in the first function expression :
$$f(0)=-frac{b}{a}=-f(a)$$
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
But is it possible to find a numerical value?
– Enzo
Nov 25 '18 at 17:17
@Enzo No, since all your given expressions involve arbitrary constants.
– Rebellos
Nov 25 '18 at 17:18
add a comment |
Well, you want the argument of your given function to be equal to zero. Thus, considering the argument of the funtion $f(ax+b) = x$, we have :
$$ax + b = 0 Leftrightarrow x = -frac{b}{a}$$
Then, $f(0)$ would be yielded for $x= -b/a$, thus again by substituting in the first function expression :
$$f(0)=-frac{b}{a}=-f(a)$$
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
Well, you want the argument of your given function to be equal to zero. Thus, considering the argument of the funtion $f(ax+b) = x$, we have :
$$ax + b = 0 Leftrightarrow x = -frac{b}{a}$$
Then, $f(0)$ would be yielded for $x= -b/a$, thus again by substituting in the first function expression :
$$f(0)=-frac{b}{a}=-f(a)$$
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
answered Nov 25 '18 at 17:17
Rebellos
14.5k31245
14.5k31245
But is it possible to find a numerical value?
– Enzo
Nov 25 '18 at 17:17
@Enzo No, since all your given expressions involve arbitrary constants.
– Rebellos
Nov 25 '18 at 17:18
add a comment |
But is it possible to find a numerical value?
– Enzo
Nov 25 '18 at 17:17
@Enzo No, since all your given expressions involve arbitrary constants.
– Rebellos
Nov 25 '18 at 17:18
But is it possible to find a numerical value?
– Enzo
Nov 25 '18 at 17:17
But is it possible to find a numerical value?
– Enzo
Nov 25 '18 at 17:17
@Enzo No, since all your given expressions involve arbitrary constants.
– Rebellos
Nov 25 '18 at 17:18
@Enzo No, since all your given expressions involve arbitrary constants.
– Rebellos
Nov 25 '18 at 17:18
add a comment |
Here is a start - and contrary to comments, there is a solution.
Let $y=ax+b$ so that $f(y)=x=cfrac {y-b}a$ so now we have expressed $f$ as a linear function of its argument.
I have not yet used the value for $f(a)$ that has been given, so this is the next step.
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
add a comment |
Here is a start - and contrary to comments, there is a solution.
Let $y=ax+b$ so that $f(y)=x=cfrac {y-b}a$ so now we have expressed $f$ as a linear function of its argument.
I have not yet used the value for $f(a)$ that has been given, so this is the next step.
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
add a comment |
Here is a start - and contrary to comments, there is a solution.
Let $y=ax+b$ so that $f(y)=x=cfrac {y-b}a$ so now we have expressed $f$ as a linear function of its argument.
I have not yet used the value for $f(a)$ that has been given, so this is the next step.
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
Here is a start - and contrary to comments, there is a solution.
Let $y=ax+b$ so that $f(y)=x=cfrac {y-b}a$ so now we have expressed $f$ as a linear function of its argument.
I have not yet used the value for $f(a)$ that has been given, so this is the next step.
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
answered Nov 25 '18 at 17:35
Mark Bennet
80.6k981179
80.6k981179
add a comment |
add a comment |
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1
Substitute $x=dfrac{-b}{a}$.
– Yadati Kiran
Nov 25 '18 at 17:14
Surely $a$ is a constant here and not a variable?
– Mark Bennet
Nov 25 '18 at 17:16
@Enzo Not possible to find a numberical value from the facts given. Also, if $a$ is a constant then using $a=0$ is wrong. You must determine $f(0)$ from the initial expression.
– Rebellos
Nov 25 '18 at 17:19
@Rebellos Assume that $a, b neq 0 in mathcal{R}$, Is it the same in this case?
– Enzo
Nov 25 '18 at 17:19
@Enzo Yes. Also if you want to typeset the set of real numbers, try mathbb R which produces $mathbb R$.
– Rebellos
Nov 25 '18 at 17:22