The following graph shows a distribution curve given by a function of the form $f(x)=A/(x+2e^{rx})$
$begingroup$
The following graph shows a distribution curve given by a function of the form $$f(x)=dfrac{A}{x+2e^{rx}}$$
Determine the equation $f(x)$ of the graph.
algebra-precalculus graphing-functions
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
asked Dec 17 '18 at 11:37
KJ26KJ26
83
$endgroup$
|
show 4 more comments
$begingroup$
The following graph shows a distribution curve given by a function of the form $$f(x)=dfrac{A}{x+2e^{rx}}$$
Determine the equation $f(x)$ of the graph.
algebra-precalculus graphing-functions
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
asked Dec 17 '18 at 11:37
KJ26KJ26
83
$endgroup$
$begingroup$
Do you mean $xe^r$ or $e^{rx}$?
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:40
$begingroup$
Welcome to the website. Please typeset your equations using Mathjax. Also, kindly include how you have tried to approach the problem so that we know you are not just trying to get your homework solved and can offer you focused guidance.
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:42
$begingroup$
Do you mean: $$f(x)=frac{A}{x+2e^{rx}}$$
$endgroup$
– projectilemotion
Dec 17 '18 at 11:52
$begingroup$
yeah @projectilemotion
$endgroup$
– KJ26
Dec 17 '18 at 11:53
1
$begingroup$
One easy way to do it is to use two of the points you were given and solve the resulting system. In my opinion, the easiest way is to first use $(0,10)$, then $(1,12)$.
$endgroup$
– projectilemotion
Dec 17 '18 at 11:55
|
show 4 more comments
$begingroup$
The following graph shows a distribution curve given by a function of the form $$f(x)=dfrac{A}{x+2e^{rx}}$$
Determine the equation $f(x)$ of the graph.
algebra-precalculus graphing-functions
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
asked Dec 17 '18 at 11:37
KJ26KJ26
83
$endgroup$
The following graph shows a distribution curve given by a function of the form $$f(x)=dfrac{A}{x+2e^{rx}}$$
Determine the equation $f(x)$ of the graph.
algebra-precalculus graphing-functions
algebra-precalculus graphing-functions
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
asked Dec 17 '18 at 11:37
KJ26KJ26
83
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
asked Dec 17 '18 at 11:37
KJ26KJ26
83
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
edited Dec 17 '18 at 11:54
projectilemotion
11.4k62141
11.4k62141
asked Dec 17 '18 at 11:37
KJ26KJ26
83
asked Dec 17 '18 at 11:37
KJ26KJ26
83
asked Dec 17 '18 at 11:37
KJ26KJ26
83
83
$begingroup$
Do you mean $xe^r$ or $e^{rx}$?
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:40
$begingroup$
Welcome to the website. Please typeset your equations using Mathjax. Also, kindly include how you have tried to approach the problem so that we know you are not just trying to get your homework solved and can offer you focused guidance.
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:42
$begingroup$
Do you mean: $$f(x)=frac{A}{x+2e^{rx}}$$
$endgroup$
– projectilemotion
Dec 17 '18 at 11:52
$begingroup$
yeah @projectilemotion
$endgroup$
– KJ26
Dec 17 '18 at 11:53
1
$begingroup$
One easy way to do it is to use two of the points you were given and solve the resulting system. In my opinion, the easiest way is to first use $(0,10)$, then $(1,12)$.
$endgroup$
– projectilemotion
Dec 17 '18 at 11:55
|
show 4 more comments
$begingroup$
Do you mean $xe^r$ or $e^{rx}$?
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:40
$begingroup$
Welcome to the website. Please typeset your equations using Mathjax. Also, kindly include how you have tried to approach the problem so that we know you are not just trying to get your homework solved and can offer you focused guidance.
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:42
$begingroup$
Do you mean: $$f(x)=frac{A}{x+2e^{rx}}$$
$endgroup$
– projectilemotion
Dec 17 '18 at 11:52
$begingroup$
yeah @projectilemotion
$endgroup$
– KJ26
Dec 17 '18 at 11:53
1
$begingroup$
One easy way to do it is to use two of the points you were given and solve the resulting system. In my opinion, the easiest way is to first use $(0,10)$, then $(1,12)$.
$endgroup$
– projectilemotion
Dec 17 '18 at 11:55
$begingroup$
Do you mean $xe^r$ or $e^{rx}$?
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:40
$begingroup$
Do you mean $xe^r$ or $e^{rx}$?
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:40
$begingroup$
Welcome to the website. Please typeset your equations using Mathjax. Also, kindly include how you have tried to approach the problem so that we know you are not just trying to get your homework solved and can offer you focused guidance.
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:42
$begingroup$
Welcome to the website. Please typeset your equations using Mathjax. Also, kindly include how you have tried to approach the problem so that we know you are not just trying to get your homework solved and can offer you focused guidance.
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:42
$begingroup$
Do you mean: $$f(x)=frac{A}{x+2e^{rx}}$$
$endgroup$
– projectilemotion
Dec 17 '18 at 11:52
$begingroup$
Do you mean: $$f(x)=frac{A}{x+2e^{rx}}$$
$endgroup$
– projectilemotion
Dec 17 '18 at 11:52
$begingroup$
yeah @projectilemotion
$endgroup$
– KJ26
Dec 17 '18 at 11:53
$begingroup$
yeah @projectilemotion
$endgroup$
– KJ26
Dec 17 '18 at 11:53
1
1
$begingroup$
One easy way to do it is to use two of the points you were given and solve the resulting system. In my opinion, the easiest way is to first use $(0,10)$, then $(1,12)$.
$endgroup$
– projectilemotion
Dec 17 '18 at 11:55
$begingroup$
One easy way to do it is to use two of the points you were given and solve the resulting system. In my opinion, the easiest way is to first use $(0,10)$, then $(1,12)$.
$endgroup$
– projectilemotion
Dec 17 '18 at 11:55
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
From the given form $f=frac{A}{1+2e^{rx}}$ and by checking the points on the graph we get
$$begin{aligned}
10&=f(0)=frac{A}{0+2e^{rcdot0}}=frac{A}{2} implies A=20 &&(text{I})\
12&=f(1)=frac{A}{1+2e^{rcdot1}}overset{text{(I)}}{=}frac{20}{1+2e^r} &&text{(II)}\
9&=f(2)=frac{A}{2+2e^{rcdot2}}overset{text{(I)}}{=}frac{10}{1+e^{2r}} &&text{(III)}\
end{aligned}$$
transform (II):
$$
12=frac{20}{1+2e^r} iff 12(1+2e^{r})=20 iff 24e^{r}=8 iff e^r=frac{1}{3} iff r=lnleft(frac{1}{3}right)
$$
So
$A=20$ and $r=ln(frac{1}{3})approx -1.0986$. (III) is not needed for this.
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
$endgroup$
add a comment |
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$begingroup$
From the given form $f=frac{A}{1+2e^{rx}}$ and by checking the points on the graph we get
$$begin{aligned}
10&=f(0)=frac{A}{0+2e^{rcdot0}}=frac{A}{2} implies A=20 &&(text{I})\
12&=f(1)=frac{A}{1+2e^{rcdot1}}overset{text{(I)}}{=}frac{20}{1+2e^r} &&text{(II)}\
9&=f(2)=frac{A}{2+2e^{rcdot2}}overset{text{(I)}}{=}frac{10}{1+e^{2r}} &&text{(III)}\
end{aligned}$$
transform (II):
$$
12=frac{20}{1+2e^r} iff 12(1+2e^{r})=20 iff 24e^{r}=8 iff e^r=frac{1}{3} iff r=lnleft(frac{1}{3}right)
$$
So
$A=20$ and $r=ln(frac{1}{3})approx -1.0986$. (III) is not needed for this.
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
$endgroup$
add a comment |
$begingroup$
From the given form $f=frac{A}{1+2e^{rx}}$ and by checking the points on the graph we get
$$begin{aligned}
10&=f(0)=frac{A}{0+2e^{rcdot0}}=frac{A}{2} implies A=20 &&(text{I})\
12&=f(1)=frac{A}{1+2e^{rcdot1}}overset{text{(I)}}{=}frac{20}{1+2e^r} &&text{(II)}\
9&=f(2)=frac{A}{2+2e^{rcdot2}}overset{text{(I)}}{=}frac{10}{1+e^{2r}} &&text{(III)}\
end{aligned}$$
transform (II):
$$
12=frac{20}{1+2e^r} iff 12(1+2e^{r})=20 iff 24e^{r}=8 iff e^r=frac{1}{3} iff r=lnleft(frac{1}{3}right)
$$
So
$A=20$ and $r=ln(frac{1}{3})approx -1.0986$. (III) is not needed for this.
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
$endgroup$
add a comment |
$begingroup$
From the given form $f=frac{A}{1+2e^{rx}}$ and by checking the points on the graph we get
$$begin{aligned}
10&=f(0)=frac{A}{0+2e^{rcdot0}}=frac{A}{2} implies A=20 &&(text{I})\
12&=f(1)=frac{A}{1+2e^{rcdot1}}overset{text{(I)}}{=}frac{20}{1+2e^r} &&text{(II)}\
9&=f(2)=frac{A}{2+2e^{rcdot2}}overset{text{(I)}}{=}frac{10}{1+e^{2r}} &&text{(III)}\
end{aligned}$$
transform (II):
$$
12=frac{20}{1+2e^r} iff 12(1+2e^{r})=20 iff 24e^{r}=8 iff e^r=frac{1}{3} iff r=lnleft(frac{1}{3}right)
$$
So
$A=20$ and $r=ln(frac{1}{3})approx -1.0986$. (III) is not needed for this.
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
$endgroup$
From the given form $f=frac{A}{1+2e^{rx}}$ and by checking the points on the graph we get
$$begin{aligned}
10&=f(0)=frac{A}{0+2e^{rcdot0}}=frac{A}{2} implies A=20 &&(text{I})\
12&=f(1)=frac{A}{1+2e^{rcdot1}}overset{text{(I)}}{=}frac{20}{1+2e^r} &&text{(II)}\
9&=f(2)=frac{A}{2+2e^{rcdot2}}overset{text{(I)}}{=}frac{10}{1+e^{2r}} &&text{(III)}\
end{aligned}$$
transform (II):
$$
12=frac{20}{1+2e^r} iff 12(1+2e^{r})=20 iff 24e^{r}=8 iff e^r=frac{1}{3} iff r=lnleft(frac{1}{3}right)
$$
So
$A=20$ and $r=ln(frac{1}{3})approx -1.0986$. (III) is not needed for this.
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
edited Dec 17 '18 at 22:29
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
answered Dec 17 '18 at 12:27
quiliupquiliup
1799
1799
add a comment |
add a comment |
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$begingroup$
Do you mean $xe^r$ or $e^{rx}$?
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:40
$begingroup$
Welcome to the website. Please typeset your equations using Mathjax. Also, kindly include how you have tried to approach the problem so that we know you are not just trying to get your homework solved and can offer you focused guidance.
$endgroup$
– Shubham Johri
Dec 17 '18 at 11:42
$begingroup$
Do you mean: $$f(x)=frac{A}{x+2e^{rx}}$$
$endgroup$
– projectilemotion
Dec 17 '18 at 11:52
$begingroup$
yeah @projectilemotion
$endgroup$
– KJ26
Dec 17 '18 at 11:53
1
$begingroup$
One easy way to do it is to use two of the points you were given and solve the resulting system. In my opinion, the easiest way is to first use $(0,10)$, then $(1,12)$.
$endgroup$
– projectilemotion
Dec 17 '18 at 11:55