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Simpler ways to show that $f(x) = x^3 - 3a^2x$ can be attained by shifting $g(x) = x^3-3ax^2$
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As the title states I'm looking for:
Simpler ways to show that graph of $f(x) = x^3 - 3a^2x$ can be attained by shifting graph of $g(x) = x^3-3ax^2$ and vice versa.
I've been recently solving another problem where i tried to find the center of symmetry for $h(x) = ax^3 + bx^2 + cx +d$. I will not post the solution for this but the result is
$$
(x_0, y_0) = (-bover3a, 2b^3 - 9abc + 27a^2dover27a^2)
$$
Since $f(x)$ is odd the center is in $(0, 0)$. With the help of the formula above one may find $(x_0, y_0)$ for $g(x)$ which is $(a,-2a^3)$. Here $a$ stands for $a$ in $g(x)$. Knowing the center of $g(x)$ we may try to shift $f(x)$ to that point and see whether $g(x)$ is obtained. So
$$
f(x-a) - 2a^3 = (x-a)^3 - 3a^2(x-a)-2a^3
$$
After expanding and canceling the terms I got $x^3 - 3ax^2$ which is equal to $g(x)$ and therefore by shifting $g(x)$ we may get $f(x)$ and vice versa.
I find this solution clumsy since I have to know that monstrous formula for the center of symmetry. Are there more elegant ways to show what the title states?
algebra-precalculus
asked Jul 18 at 17:06
roman
4391413
 |Â
show 1 more comment
up vote
0
down vote
favorite
As the title states I'm looking for:
Simpler ways to show that graph of $f(x) = x^3 - 3a^2x$ can be attained by shifting graph of $g(x) = x^3-3ax^2$ and vice versa.
I've been recently solving another problem where i tried to find the center of symmetry for $h(x) = ax^3 + bx^2 + cx +d$. I will not post the solution for this but the result is
$$
(x_0, y_0) = (-bover3a, 2b^3 - 9abc + 27a^2dover27a^2)
$$
Since $f(x)$ is odd the center is in $(0, 0)$. With the help of the formula above one may find $(x_0, y_0)$ for $g(x)$ which is $(a,-2a^3)$. Here $a$ stands for $a$ in $g(x)$. Knowing the center of $g(x)$ we may try to shift $f(x)$ to that point and see whether $g(x)$ is obtained. So
$$
f(x-a) - 2a^3 = (x-a)^3 - 3a^2(x-a)-2a^3
$$
After expanding and canceling the terms I got $x^3 - 3ax^2$ which is equal to $g(x)$ and therefore by shifting $g(x)$ we may get $f(x)$ and vice versa.
I find this solution clumsy since I have to know that monstrous formula for the center of symmetry. Are there more elegant ways to show what the title states?
algebra-precalculus
asked Jul 18 at 17:06
roman
4391413
2
An almost trivial way: we want to eliminate the $x^2$ term, so consider $g(x+t)$, for some $t$ to be determined. Expand using the binomial theorem and find the value of $t$ that makes the $x^2$ term zero.
– Steven Stadnicki
Jul 18 at 17:13
@StevenStadnicki I've just tried what you suggest but I am not really getting how one could eliminate $x^2$. The roots of $f(x+t)$ are $-x$ and $3a-x$ but they turn the whole function into zero.
– roman
Jul 18 at 17:39
$g(x+t) = (x+t)^3-3a(x+t)^2$ $= x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2$ $=ldots$
– Steven Stadnicki
Jul 18 at 17:41
@StevenStadnicki ok, so when $t = a$ i get $x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2 = x^3 - 3a^2x-2a^3$ and this eliminates the $x^2$ term, but that still doesn't give me insights. $f(x+a) = x^3 + 3x^2a - 2a^3$ and $g(x+a) = x^3 - 3a^2x - 2a^3$
– roman
Jul 18 at 17:55
Why would you want to compute $f(x+a)$?
– amd
Jul 18 at 18:35
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As the title states I'm looking for:
Simpler ways to show that graph of $f(x) = x^3 - 3a^2x$ can be attained by shifting graph of $g(x) = x^3-3ax^2$ and vice versa.
I've been recently solving another problem where i tried to find the center of symmetry for $h(x) = ax^3 + bx^2 + cx +d$. I will not post the solution for this but the result is
$$
(x_0, y_0) = (-bover3a, 2b^3 - 9abc + 27a^2dover27a^2)
$$
Since $f(x)$ is odd the center is in $(0, 0)$. With the help of the formula above one may find $(x_0, y_0)$ for $g(x)$ which is $(a,-2a^3)$. Here $a$ stands for $a$ in $g(x)$. Knowing the center of $g(x)$ we may try to shift $f(x)$ to that point and see whether $g(x)$ is obtained. So
$$
f(x-a) - 2a^3 = (x-a)^3 - 3a^2(x-a)-2a^3
$$
After expanding and canceling the terms I got $x^3 - 3ax^2$ which is equal to $g(x)$ and therefore by shifting $g(x)$ we may get $f(x)$ and vice versa.
I find this solution clumsy since I have to know that monstrous formula for the center of symmetry. Are there more elegant ways to show what the title states?
algebra-precalculus
asked Jul 18 at 17:06
roman
4391413
As the title states I'm looking for:
Simpler ways to show that graph of $f(x) = x^3 - 3a^2x$ can be attained by shifting graph of $g(x) = x^3-3ax^2$ and vice versa.
I've been recently solving another problem where i tried to find the center of symmetry for $h(x) = ax^3 + bx^2 + cx +d$. I will not post the solution for this but the result is
$$
(x_0, y_0) = (-bover3a, 2b^3 - 9abc + 27a^2dover27a^2)
$$
Since $f(x)$ is odd the center is in $(0, 0)$. With the help of the formula above one may find $(x_0, y_0)$ for $g(x)$ which is $(a,-2a^3)$. Here $a$ stands for $a$ in $g(x)$. Knowing the center of $g(x)$ we may try to shift $f(x)$ to that point and see whether $g(x)$ is obtained. So
$$
f(x-a) - 2a^3 = (x-a)^3 - 3a^2(x-a)-2a^3
$$
After expanding and canceling the terms I got $x^3 - 3ax^2$ which is equal to $g(x)$ and therefore by shifting $g(x)$ we may get $f(x)$ and vice versa.
I find this solution clumsy since I have to know that monstrous formula for the center of symmetry. Are there more elegant ways to show what the title states?
algebra-precalculus
asked Jul 18 at 17:06
roman
4391413
asked Jul 18 at 17:06
roman
4391413
asked Jul 18 at 17:06
roman
4391413
asked Jul 18 at 17:06
roman
4391413
4391413
2
An almost trivial way: we want to eliminate the $x^2$ term, so consider $g(x+t)$, for some $t$ to be determined. Expand using the binomial theorem and find the value of $t$ that makes the $x^2$ term zero.
– Steven Stadnicki
Jul 18 at 17:13
@StevenStadnicki I've just tried what you suggest but I am not really getting how one could eliminate $x^2$. The roots of $f(x+t)$ are $-x$ and $3a-x$ but they turn the whole function into zero.
– roman
Jul 18 at 17:39
$g(x+t) = (x+t)^3-3a(x+t)^2$ $= x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2$ $=ldots$
– Steven Stadnicki
Jul 18 at 17:41
@StevenStadnicki ok, so when $t = a$ i get $x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2 = x^3 - 3a^2x-2a^3$ and this eliminates the $x^2$ term, but that still doesn't give me insights. $f(x+a) = x^3 + 3x^2a - 2a^3$ and $g(x+a) = x^3 - 3a^2x - 2a^3$
– roman
Jul 18 at 17:55
Why would you want to compute $f(x+a)$?
– amd
Jul 18 at 18:35
 |Â
show 1 more comment
2
An almost trivial way: we want to eliminate the $x^2$ term, so consider $g(x+t)$, for some $t$ to be determined. Expand using the binomial theorem and find the value of $t$ that makes the $x^2$ term zero.
– Steven Stadnicki
Jul 18 at 17:13
@StevenStadnicki I've just tried what you suggest but I am not really getting how one could eliminate $x^2$. The roots of $f(x+t)$ are $-x$ and $3a-x$ but they turn the whole function into zero.
– roman
Jul 18 at 17:39
$g(x+t) = (x+t)^3-3a(x+t)^2$ $= x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2$ $=ldots$
– Steven Stadnicki
Jul 18 at 17:41
@StevenStadnicki ok, so when $t = a$ i get $x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2 = x^3 - 3a^2x-2a^3$ and this eliminates the $x^2$ term, but that still doesn't give me insights. $f(x+a) = x^3 + 3x^2a - 2a^3$ and $g(x+a) = x^3 - 3a^2x - 2a^3$
– roman
Jul 18 at 17:55
Why would you want to compute $f(x+a)$?
– amd
Jul 18 at 18:35
2
2
An almost trivial way: we want to eliminate the $x^2$ term, so consider $g(x+t)$, for some $t$ to be determined. Expand using the binomial theorem and find the value of $t$ that makes the $x^2$ term zero.
– Steven Stadnicki
Jul 18 at 17:13
An almost trivial way: we want to eliminate the $x^2$ term, so consider $g(x+t)$, for some $t$ to be determined. Expand using the binomial theorem and find the value of $t$ that makes the $x^2$ term zero.
– Steven Stadnicki
Jul 18 at 17:13
@StevenStadnicki I've just tried what you suggest but I am not really getting how one could eliminate $x^2$. The roots of $f(x+t)$ are $-x$ and $3a-x$ but they turn the whole function into zero.
– roman
Jul 18 at 17:39
@StevenStadnicki I've just tried what you suggest but I am not really getting how one could eliminate $x^2$. The roots of $f(x+t)$ are $-x$ and $3a-x$ but they turn the whole function into zero.
– roman
Jul 18 at 17:39
$g(x+t) = (x+t)^3-3a(x+t)^2$ $= x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2$ $=ldots$
– Steven Stadnicki
Jul 18 at 17:41
$g(x+t) = (x+t)^3-3a(x+t)^2$ $= x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2$ $=ldots$
– Steven Stadnicki
Jul 18 at 17:41
@StevenStadnicki ok, so when $t = a$ i get $x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2 = x^3 - 3a^2x-2a^3$ and this eliminates the $x^2$ term, but that still doesn't give me insights. $f(x+a) = x^3 + 3x^2a - 2a^3$ and $g(x+a) = x^3 - 3a^2x - 2a^3$
– roman
Jul 18 at 17:55
@StevenStadnicki ok, so when $t = a$ i get $x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2 = x^3 - 3a^2x-2a^3$ and this eliminates the $x^2$ term, but that still doesn't give me insights. $f(x+a) = x^3 + 3x^2a - 2a^3$ and $g(x+a) = x^3 - 3a^2x - 2a^3$
– roman
Jul 18 at 17:55
Why would you want to compute $f(x+a)$?
– amd
Jul 18 at 18:35
Why would you want to compute $f(x+a)$?
– amd
Jul 18 at 18:35
 |Â
show 1 more comment
2 Answers
2
active
oldest
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up vote
1
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Following Steven Stadnicki’s comment, expand $g(x-h)+k$, which is $g$ shifted right by $h$ and up by $k$, with negative values shifting in the opposite directions: $$g(x-h)+k = x^3-3(h+a)x^2+3h(h+2a)x+(k-3ah^2-h^3).$$ You want the $x^2$ term to vanish, so $h=-a$. Substituting this value of $h$ into the above expression, you then have $$x^3-3a^2x+(k-2a^3)$$ from which the correct value of $k$ should also be obvious. The graph of $g$ is then of course obtained from that of $f$ by making the opposite shift.
You could also try a similar approach by starting with $f(x-h)+k$ and finding an $h$ that eliminates the $x$ term, but that’s going to be a bit more work: there are two possibilities, one of which you’ll end up discarding.
answered Jul 18 at 18:45
amd
25.9k2943
add a comment |Â
up vote
1
down vote
If $f(x)$ and $g(x)$ are the shifted versions, so must be their 1st-order derivations. We have $$f'(x)=3(x^2-a^2)\g'(x)=3(x^2-2ax)$$it's clear that $$f'(x-a)=g'(x)$$therefore by integrating $$f(x-a)=g(x)+C$$also $$f(0)=g(a)+C$$therefore $$C=2a^3$$ if $f(x)$ and $g(x)$ are shifted versions we should have $C=0$ or $a=0$
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Following Steven Stadnicki’s comment, expand $g(x-h)+k$, which is $g$ shifted right by $h$ and up by $k$, with negative values shifting in the opposite directions: $$g(x-h)+k = x^3-3(h+a)x^2+3h(h+2a)x+(k-3ah^2-h^3).$$ You want the $x^2$ term to vanish, so $h=-a$. Substituting this value of $h$ into the above expression, you then have $$x^3-3a^2x+(k-2a^3)$$ from which the correct value of $k$ should also be obvious. The graph of $g$ is then of course obtained from that of $f$ by making the opposite shift.
You could also try a similar approach by starting with $f(x-h)+k$ and finding an $h$ that eliminates the $x$ term, but that’s going to be a bit more work: there are two possibilities, one of which you’ll end up discarding.
answered Jul 18 at 18:45
amd
25.9k2943
add a comment |Â
up vote
1
down vote
accepted
Following Steven Stadnicki’s comment, expand $g(x-h)+k$, which is $g$ shifted right by $h$ and up by $k$, with negative values shifting in the opposite directions: $$g(x-h)+k = x^3-3(h+a)x^2+3h(h+2a)x+(k-3ah^2-h^3).$$ You want the $x^2$ term to vanish, so $h=-a$. Substituting this value of $h$ into the above expression, you then have $$x^3-3a^2x+(k-2a^3)$$ from which the correct value of $k$ should also be obvious. The graph of $g$ is then of course obtained from that of $f$ by making the opposite shift.
You could also try a similar approach by starting with $f(x-h)+k$ and finding an $h$ that eliminates the $x$ term, but that’s going to be a bit more work: there are two possibilities, one of which you’ll end up discarding.
answered Jul 18 at 18:45
amd
25.9k2943
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Following Steven Stadnicki’s comment, expand $g(x-h)+k$, which is $g$ shifted right by $h$ and up by $k$, with negative values shifting in the opposite directions: $$g(x-h)+k = x^3-3(h+a)x^2+3h(h+2a)x+(k-3ah^2-h^3).$$ You want the $x^2$ term to vanish, so $h=-a$. Substituting this value of $h$ into the above expression, you then have $$x^3-3a^2x+(k-2a^3)$$ from which the correct value of $k$ should also be obvious. The graph of $g$ is then of course obtained from that of $f$ by making the opposite shift.
You could also try a similar approach by starting with $f(x-h)+k$ and finding an $h$ that eliminates the $x$ term, but that’s going to be a bit more work: there are two possibilities, one of which you’ll end up discarding.
answered Jul 18 at 18:45
amd
25.9k2943
Following Steven Stadnicki’s comment, expand $g(x-h)+k$, which is $g$ shifted right by $h$ and up by $k$, with negative values shifting in the opposite directions: $$g(x-h)+k = x^3-3(h+a)x^2+3h(h+2a)x+(k-3ah^2-h^3).$$ You want the $x^2$ term to vanish, so $h=-a$. Substituting this value of $h$ into the above expression, you then have $$x^3-3a^2x+(k-2a^3)$$ from which the correct value of $k$ should also be obvious. The graph of $g$ is then of course obtained from that of $f$ by making the opposite shift.
You could also try a similar approach by starting with $f(x-h)+k$ and finding an $h$ that eliminates the $x$ term, but that’s going to be a bit more work: there are two possibilities, one of which you’ll end up discarding.
answered Jul 18 at 18:45
amd
25.9k2943
edited Jul 19 at 19:20
answered Jul 18 at 18:45
amd
25.9k2943
answered Jul 18 at 18:45
amd
25.9k2943
answered Jul 18 at 18:45
amd
25.9k2943
25.9k2943
add a comment |Â
add a comment |Â
up vote
1
down vote
If $f(x)$ and $g(x)$ are the shifted versions, so must be their 1st-order derivations. We have $$f'(x)=3(x^2-a^2)\g'(x)=3(x^2-2ax)$$it's clear that $$f'(x-a)=g'(x)$$therefore by integrating $$f(x-a)=g(x)+C$$also $$f(0)=g(a)+C$$therefore $$C=2a^3$$ if $f(x)$ and $g(x)$ are shifted versions we should have $C=0$ or $a=0$
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
If $f(x)$ and $g(x)$ are the shifted versions, so must be their 1st-order derivations. We have $$f'(x)=3(x^2-a^2)\g'(x)=3(x^2-2ax)$$it's clear that $$f'(x-a)=g'(x)$$therefore by integrating $$f(x-a)=g(x)+C$$also $$f(0)=g(a)+C$$therefore $$C=2a^3$$ if $f(x)$ and $g(x)$ are shifted versions we should have $C=0$ or $a=0$
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $f(x)$ and $g(x)$ are the shifted versions, so must be their 1st-order derivations. We have $$f'(x)=3(x^2-a^2)\g'(x)=3(x^2-2ax)$$it's clear that $$f'(x-a)=g'(x)$$therefore by integrating $$f(x-a)=g(x)+C$$also $$f(0)=g(a)+C$$therefore $$C=2a^3$$ if $f(x)$ and $g(x)$ are shifted versions we should have $C=0$ or $a=0$
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
If $f(x)$ and $g(x)$ are the shifted versions, so must be their 1st-order derivations. We have $$f'(x)=3(x^2-a^2)\g'(x)=3(x^2-2ax)$$it's clear that $$f'(x-a)=g'(x)$$therefore by integrating $$f(x-a)=g(x)+C$$also $$f(0)=g(a)+C$$therefore $$C=2a^3$$ if $f(x)$ and $g(x)$ are shifted versions we should have $C=0$ or $a=0$
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
answered Jul 18 at 18:07
Mostafa Ayaz
8,6023630
8,6023630
add a comment |Â
add a comment |Â
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2
An almost trivial way: we want to eliminate the $x^2$ term, so consider $g(x+t)$, for some $t$ to be determined. Expand using the binomial theorem and find the value of $t$ that makes the $x^2$ term zero.
– Steven Stadnicki
Jul 18 at 17:13
@StevenStadnicki I've just tried what you suggest but I am not really getting how one could eliminate $x^2$. The roots of $f(x+t)$ are $-x$ and $3a-x$ but they turn the whole function into zero.
– roman
Jul 18 at 17:39
$g(x+t) = (x+t)^3-3a(x+t)^2$ $= x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2$ $=ldots$
– Steven Stadnicki
Jul 18 at 17:41
@StevenStadnicki ok, so when $t = a$ i get $x^3+3tx^2+3t^2x+t^3-3ax^2-6atx-3at^2 = x^3 - 3a^2x-2a^3$ and this eliminates the $x^2$ term, but that still doesn't give me insights. $f(x+a) = x^3 + 3x^2a - 2a^3$ and $g(x+a) = x^3 - 3a^2x - 2a^3$
– roman
Jul 18 at 17:55
Why would you want to compute $f(x+a)$?
– amd
Jul 18 at 18:35