Mixing
Showing properties of a surge function
$begingroup$
I am working on the question below and I am getting stuck.
Consider the surge function $y=axe^{-bx}$ with $a$ and $b$ positive constants.
(a) Find the local maxima, local minima, and inflection points.
(b) How does varying $a$ and $b$ affect the shape of the graph?
(c) On one set of axes, sketch the graph of this function for a few values of $a$ and $b$.
I've played around with the function on Wolfram Alpha in order to get a feel for how different values of $a$ and $b$ affect the graph. I also see that the local maximum seems always to be at $1/b$. I cant seem to figure out how to show this result using first and second derivatives, etc.
I've found the first derivative to be: $$y'=ae^{-bx}(1-b)$$
calculus
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
$endgroup$
add a comment |
$begingroup$
I am working on the question below and I am getting stuck.
Consider the surge function $y=axe^{-bx}$ with $a$ and $b$ positive constants.
(a) Find the local maxima, local minima, and inflection points.
(b) How does varying $a$ and $b$ affect the shape of the graph?
(c) On one set of axes, sketch the graph of this function for a few values of $a$ and $b$.
I've played around with the function on Wolfram Alpha in order to get a feel for how different values of $a$ and $b$ affect the graph. I also see that the local maximum seems always to be at $1/b$. I cant seem to figure out how to show this result using first and second derivatives, etc.
I've found the first derivative to be: $$y'=ae^{-bx}(1-b)$$
calculus
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
$endgroup$
4
$begingroup$
$y'=ae^{-bx}-baxe^{-bx} = ae^{-bx}(1-bx)$.
$endgroup$
– David Mitra
Nov 17 '11 at 8:27
$begingroup$
Oops, yes I see that now, I had not factored it properly. So $(1-bx)=0$, which leads to $x=1/b$.
$endgroup$
– NehriMattise
Nov 17 '11 at 8:57
$begingroup$
Assuming $a>0$ and $b>0$ then all the possible curves are similar. You have $left(frac b aright) y = (bx)e^{-bx}$, so increasing $a$ while leaving $b$ unchanged stretches the curve away from the $x$-axis (i.e. increases the magnitude of $y$ while leaving $x$ unchanged), while increasing $a$ and $b$ by the same proportionate amount shrinks the curve towards the $y$-axis (i.e. reduces the magnitude of $x$ while leaving $x$ unchanged). So increasing $b$ without changing $a$ would shrink towards the origin. Negative $a$ or $b$ would just flip the curve across the axes
$endgroup$
– Henry
Jul 25 '18 at 12:22
add a comment |
$begingroup$
I am working on the question below and I am getting stuck.
Consider the surge function $y=axe^{-bx}$ with $a$ and $b$ positive constants.
(a) Find the local maxima, local minima, and inflection points.
(b) How does varying $a$ and $b$ affect the shape of the graph?
(c) On one set of axes, sketch the graph of this function for a few values of $a$ and $b$.
I've played around with the function on Wolfram Alpha in order to get a feel for how different values of $a$ and $b$ affect the graph. I also see that the local maximum seems always to be at $1/b$. I cant seem to figure out how to show this result using first and second derivatives, etc.
I've found the first derivative to be: $$y'=ae^{-bx}(1-b)$$
calculus
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
$endgroup$
I am working on the question below and I am getting stuck.
Consider the surge function $y=axe^{-bx}$ with $a$ and $b$ positive constants.
(a) Find the local maxima, local minima, and inflection points.
(b) How does varying $a$ and $b$ affect the shape of the graph?
(c) On one set of axes, sketch the graph of this function for a few values of $a$ and $b$.
I've played around with the function on Wolfram Alpha in order to get a feel for how different values of $a$ and $b$ affect the graph. I also see that the local maximum seems always to be at $1/b$. I cant seem to figure out how to show this result using first and second derivatives, etc.
I've found the first derivative to be: $$y'=ae^{-bx}(1-b)$$
calculus
calculus
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
asked Nov 17 '11 at 8:22
NehriMattiseNehriMattise
55124
55124
4
$begingroup$
$y'=ae^{-bx}-baxe^{-bx} = ae^{-bx}(1-bx)$.
$endgroup$
– David Mitra
Nov 17 '11 at 8:27
$begingroup$
Oops, yes I see that now, I had not factored it properly. So $(1-bx)=0$, which leads to $x=1/b$.
$endgroup$
– NehriMattise
Nov 17 '11 at 8:57
$begingroup$
Assuming $a>0$ and $b>0$ then all the possible curves are similar. You have $left(frac b aright) y = (bx)e^{-bx}$, so increasing $a$ while leaving $b$ unchanged stretches the curve away from the $x$-axis (i.e. increases the magnitude of $y$ while leaving $x$ unchanged), while increasing $a$ and $b$ by the same proportionate amount shrinks the curve towards the $y$-axis (i.e. reduces the magnitude of $x$ while leaving $x$ unchanged). So increasing $b$ without changing $a$ would shrink towards the origin. Negative $a$ or $b$ would just flip the curve across the axes
$endgroup$
– Henry
Jul 25 '18 at 12:22
add a comment |
4
$begingroup$
$y'=ae^{-bx}-baxe^{-bx} = ae^{-bx}(1-bx)$.
$endgroup$
– David Mitra
Nov 17 '11 at 8:27
$begingroup$
Oops, yes I see that now, I had not factored it properly. So $(1-bx)=0$, which leads to $x=1/b$.
$endgroup$
– NehriMattise
Nov 17 '11 at 8:57
$begingroup$
Assuming $a>0$ and $b>0$ then all the possible curves are similar. You have $left(frac b aright) y = (bx)e^{-bx}$, so increasing $a$ while leaving $b$ unchanged stretches the curve away from the $x$-axis (i.e. increases the magnitude of $y$ while leaving $x$ unchanged), while increasing $a$ and $b$ by the same proportionate amount shrinks the curve towards the $y$-axis (i.e. reduces the magnitude of $x$ while leaving $x$ unchanged). So increasing $b$ without changing $a$ would shrink towards the origin. Negative $a$ or $b$ would just flip the curve across the axes
$endgroup$
– Henry
Jul 25 '18 at 12:22
4
4
$begingroup$
$y'=ae^{-bx}-baxe^{-bx} = ae^{-bx}(1-bx)$.
$endgroup$
– David Mitra
Nov 17 '11 at 8:27
$begingroup$
$y'=ae^{-bx}-baxe^{-bx} = ae^{-bx}(1-bx)$.
$endgroup$
– David Mitra
Nov 17 '11 at 8:27
$begingroup$
Oops, yes I see that now, I had not factored it properly. So $(1-bx)=0$, which leads to $x=1/b$.
$endgroup$
– NehriMattise
Nov 17 '11 at 8:57
$begingroup$
Oops, yes I see that now, I had not factored it properly. So $(1-bx)=0$, which leads to $x=1/b$.
$endgroup$
– NehriMattise
Nov 17 '11 at 8:57
$begingroup$
Assuming $a>0$ and $b>0$ then all the possible curves are similar. You have $left(frac b aright) y = (bx)e^{-bx}$, so increasing $a$ while leaving $b$ unchanged stretches the curve away from the $x$-axis (i.e. increases the magnitude of $y$ while leaving $x$ unchanged), while increasing $a$ and $b$ by the same proportionate amount shrinks the curve towards the $y$-axis (i.e. reduces the magnitude of $x$ while leaving $x$ unchanged). So increasing $b$ without changing $a$ would shrink towards the origin. Negative $a$ or $b$ would just flip the curve across the axes
$endgroup$
– Henry
Jul 25 '18 at 12:22
$begingroup$
Assuming $a>0$ and $b>0$ then all the possible curves are similar. You have $left(frac b aright) y = (bx)e^{-bx}$, so increasing $a$ while leaving $b$ unchanged stretches the curve away from the $x$-axis (i.e. increases the magnitude of $y$ while leaving $x$ unchanged), while increasing $a$ and $b$ by the same proportionate amount shrinks the curve towards the $y$-axis (i.e. reduces the magnitude of $x$ while leaving $x$ unchanged). So increasing $b$ without changing $a$ would shrink towards the origin. Negative $a$ or $b$ would just flip the curve across the axes
$endgroup$
– Henry
Jul 25 '18 at 12:22
add a comment |
1 Answer
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$begingroup$
From David Mitra:
$$y^prime =ae^{−bx}−baxe^{−bx}=ae^{−bx}(1−bx)$$
From NehriMattisse:
So $(1−bx)=0$, which leads to $x=1/b$.
$endgroup$
add a comment |
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$begingroup$
From David Mitra:
$$y^prime =ae^{−bx}−baxe^{−bx}=ae^{−bx}(1−bx)$$
From NehriMattisse:
So $(1−bx)=0$, which leads to $x=1/b$.
$endgroup$
add a comment |
$begingroup$
From David Mitra:
$$y^prime =ae^{−bx}−baxe^{−bx}=ae^{−bx}(1−bx)$$
From NehriMattisse:
So $(1−bx)=0$, which leads to $x=1/b$.
$endgroup$
add a comment |
$begingroup$
From David Mitra:
$$y^prime =ae^{−bx}−baxe^{−bx}=ae^{−bx}(1−bx)$$
From NehriMattisse:
So $(1−bx)=0$, which leads to $x=1/b$.
$endgroup$
From David Mitra:
$$y^prime =ae^{−bx}−baxe^{−bx}=ae^{−bx}(1−bx)$$
From NehriMattisse:
So $(1−bx)=0$, which leads to $x=1/b$.
answered May 19 '14 at 2:23
community wiki
Austin Mohr
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4
$begingroup$
$y'=ae^{-bx}-baxe^{-bx} = ae^{-bx}(1-bx)$.
$endgroup$
– David Mitra
Nov 17 '11 at 8:27
$begingroup$
Oops, yes I see that now, I had not factored it properly. So $(1-bx)=0$, which leads to $x=1/b$.
$endgroup$
– NehriMattise
Nov 17 '11 at 8:57
$begingroup$
Assuming $a>0$ and $b>0$ then all the possible curves are similar. You have $left(frac b aright) y = (bx)e^{-bx}$, so increasing $a$ while leaving $b$ unchanged stretches the curve away from the $x$-axis (i.e. increases the magnitude of $y$ while leaving $x$ unchanged), while increasing $a$ and $b$ by the same proportionate amount shrinks the curve towards the $y$-axis (i.e. reduces the magnitude of $x$ while leaving $x$ unchanged). So increasing $b$ without changing $a$ would shrink towards the origin. Negative $a$ or $b$ would just flip the curve across the axes
$endgroup$
– Henry
Jul 25 '18 at 12:22