Mixing
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Increasing and decreasing intervals
Clash Royale CLAN TAG#URR8PPP
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Find a polynomial $f(x)$ of degree 4 which increases in the intervals $(-∞,1)$ and $(2,3)$ and decreases in the intervals $(1,2)$ and $(3,∞)$ and satisfies the condition $f(0)=1$
It is evident that the function should be $f(x)=ax^4+bx^3+cx^2+dx+1$.
I differentiated it. Now, I'm lost.
I tried putting $f'(1)>0$, $f'(3)-f'(2)geq 0$, and $f'(2)-f'(1)leq 0$. Am I doing correct? Or is there another method?
functions derivatives polynomials
edited yesterday
greedoid
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asked yesterday
Arka Seth
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add a comment |Â
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favorite
Find a polynomial $f(x)$ of degree 4 which increases in the intervals $(-∞,1)$ and $(2,3)$ and decreases in the intervals $(1,2)$ and $(3,∞)$ and satisfies the condition $f(0)=1$
It is evident that the function should be $f(x)=ax^4+bx^3+cx^2+dx+1$.
I differentiated it. Now, I'm lost.
I tried putting $f'(1)>0$, $f'(3)-f'(2)geq 0$, and $f'(2)-f'(1)leq 0$. Am I doing correct? Or is there another method?
functions derivatives polynomials
edited yesterday
greedoid
26.1k93373
asked yesterday
Arka Seth
73
set 1, 2, and 3 as local maximum or minimum.
– user295959
yesterday
@user295959 thanks for the help!
– Arka Seth
yesterday
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find a polynomial $f(x)$ of degree 4 which increases in the intervals $(-∞,1)$ and $(2,3)$ and decreases in the intervals $(1,2)$ and $(3,∞)$ and satisfies the condition $f(0)=1$
It is evident that the function should be $f(x)=ax^4+bx^3+cx^2+dx+1$.
I differentiated it. Now, I'm lost.
I tried putting $f'(1)>0$, $f'(3)-f'(2)geq 0$, and $f'(2)-f'(1)leq 0$. Am I doing correct? Or is there another method?
functions derivatives polynomials
edited yesterday
greedoid
26.1k93373
asked yesterday
Arka Seth
73
Find a polynomial $f(x)$ of degree 4 which increases in the intervals $(-∞,1)$ and $(2,3)$ and decreases in the intervals $(1,2)$ and $(3,∞)$ and satisfies the condition $f(0)=1$
It is evident that the function should be $f(x)=ax^4+bx^3+cx^2+dx+1$.
I differentiated it. Now, I'm lost.
I tried putting $f'(1)>0$, $f'(3)-f'(2)geq 0$, and $f'(2)-f'(1)leq 0$. Am I doing correct? Or is there another method?
functions derivatives polynomials
edited yesterday
greedoid
26.1k93373
asked yesterday
Arka Seth
73
edited yesterday
greedoid
26.1k93373
edited yesterday
greedoid
26.1k93373
edited yesterday
greedoid
26.1k93373
26.1k93373
asked yesterday
Arka Seth
73
asked yesterday
Arka Seth
73
asked yesterday
Arka Seth
73
73
set 1, 2, and 3 as local maximum or minimum.
– user295959
yesterday
@user295959 thanks for the help!
– Arka Seth
yesterday
add a comment |Â
set 1, 2, and 3 as local maximum or minimum.
– user295959
yesterday
@user295959 thanks for the help!
– Arka Seth
yesterday
set 1, 2, and 3 as local maximum or minimum.
– user295959
yesterday
set 1, 2, and 3 as local maximum or minimum.
– user295959
yesterday
@user295959 thanks for the help!
– Arka Seth
yesterday
@user295959 thanks for the help!
– Arka Seth
yesterday
add a comment |Â
2 Answers
2
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1
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According to the given conditions, we may assume $$f(x)=ax^4+bx^3+cx^2+dx+1.$$
Then $$f'(x)=4ax^3+3bx^2+2cx+d.$$ Notice that $x=1,2,3$ are the roots for $f'(x)=0$. Hence
$$f'(x)=4ax^3+3bx^2+2cx+d=k(x-1)(x-2)(x-3)=kx^3-6kx^2+11kx-6k.$$
Thus, $$a=frack4,~~~b=-2k,~~~c=frac112k,~~~d=-6k.$$
Therefore, we obtain $$f(x)=frack4x^4-2kx^3+frac11k2x^2-6kx+1,$$ where $k>0.$
answered yesterday
mengdie1982
2,780216
Thanks for the help!
– Arka Seth
yesterday
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0
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Since $1,2,3$ are zeroes for $f'$ we have $$ f'(x) = a(x-1)(x-2)(x-3) = ax^3-6ax^2+11ax-6a$$
So, by integrating we get $$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +c$$
Puting $f(0)=1$ we get $c=1$ and thus:$$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +1$$
answered yesterday
greedoid
26.1k93373
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
According to the given conditions, we may assume $$f(x)=ax^4+bx^3+cx^2+dx+1.$$
Then $$f'(x)=4ax^3+3bx^2+2cx+d.$$ Notice that $x=1,2,3$ are the roots for $f'(x)=0$. Hence
$$f'(x)=4ax^3+3bx^2+2cx+d=k(x-1)(x-2)(x-3)=kx^3-6kx^2+11kx-6k.$$
Thus, $$a=frack4,~~~b=-2k,~~~c=frac112k,~~~d=-6k.$$
Therefore, we obtain $$f(x)=frack4x^4-2kx^3+frac11k2x^2-6kx+1,$$ where $k>0.$
answered yesterday
mengdie1982
2,780216
Thanks for the help!
– Arka Seth
yesterday
add a comment |Â
up vote
1
down vote
accepted
According to the given conditions, we may assume $$f(x)=ax^4+bx^3+cx^2+dx+1.$$
Then $$f'(x)=4ax^3+3bx^2+2cx+d.$$ Notice that $x=1,2,3$ are the roots for $f'(x)=0$. Hence
$$f'(x)=4ax^3+3bx^2+2cx+d=k(x-1)(x-2)(x-3)=kx^3-6kx^2+11kx-6k.$$
Thus, $$a=frack4,~~~b=-2k,~~~c=frac112k,~~~d=-6k.$$
Therefore, we obtain $$f(x)=frack4x^4-2kx^3+frac11k2x^2-6kx+1,$$ where $k>0.$
answered yesterday
mengdie1982
2,780216
Thanks for the help!
– Arka Seth
yesterday
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
According to the given conditions, we may assume $$f(x)=ax^4+bx^3+cx^2+dx+1.$$
Then $$f'(x)=4ax^3+3bx^2+2cx+d.$$ Notice that $x=1,2,3$ are the roots for $f'(x)=0$. Hence
$$f'(x)=4ax^3+3bx^2+2cx+d=k(x-1)(x-2)(x-3)=kx^3-6kx^2+11kx-6k.$$
Thus, $$a=frack4,~~~b=-2k,~~~c=frac112k,~~~d=-6k.$$
Therefore, we obtain $$f(x)=frack4x^4-2kx^3+frac11k2x^2-6kx+1,$$ where $k>0.$
answered yesterday
mengdie1982
2,780216
According to the given conditions, we may assume $$f(x)=ax^4+bx^3+cx^2+dx+1.$$
Then $$f'(x)=4ax^3+3bx^2+2cx+d.$$ Notice that $x=1,2,3$ are the roots for $f'(x)=0$. Hence
$$f'(x)=4ax^3+3bx^2+2cx+d=k(x-1)(x-2)(x-3)=kx^3-6kx^2+11kx-6k.$$
Thus, $$a=frack4,~~~b=-2k,~~~c=frac112k,~~~d=-6k.$$
Therefore, we obtain $$f(x)=frack4x^4-2kx^3+frac11k2x^2-6kx+1,$$ where $k>0.$
answered yesterday
mengdie1982
2,780216
edited yesterday
answered yesterday
mengdie1982
2,780216
answered yesterday
mengdie1982
2,780216
answered yesterday
mengdie1982
2,780216
2,780216
Thanks for the help!
– Arka Seth
yesterday
add a comment |Â
Thanks for the help!
– Arka Seth
yesterday
Thanks for the help!
– Arka Seth
yesterday
Thanks for the help!
– Arka Seth
yesterday
add a comment |Â
up vote
0
down vote
Since $1,2,3$ are zeroes for $f'$ we have $$ f'(x) = a(x-1)(x-2)(x-3) = ax^3-6ax^2+11ax-6a$$
So, by integrating we get $$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +c$$
Puting $f(0)=1$ we get $c=1$ and thus:$$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +1$$
answered yesterday
greedoid
26.1k93373
add a comment |Â
up vote
0
down vote
Since $1,2,3$ are zeroes for $f'$ we have $$ f'(x) = a(x-1)(x-2)(x-3) = ax^3-6ax^2+11ax-6a$$
So, by integrating we get $$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +c$$
Puting $f(0)=1$ we get $c=1$ and thus:$$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +1$$
answered yesterday
greedoid
26.1k93373
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $1,2,3$ are zeroes for $f'$ we have $$ f'(x) = a(x-1)(x-2)(x-3) = ax^3-6ax^2+11ax-6a$$
So, by integrating we get $$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +c$$
Puting $f(0)=1$ we get $c=1$ and thus:$$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +1$$
answered yesterday
greedoid
26.1k93373
Since $1,2,3$ are zeroes for $f'$ we have $$ f'(x) = a(x-1)(x-2)(x-3) = ax^3-6ax^2+11ax-6a$$
So, by integrating we get $$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +c$$
Puting $f(0)=1$ we get $c=1$ and thus:$$f(x) = aover 4x^4 -2ax^3-11aover 2x^2-6ax +1$$
answered yesterday
greedoid
26.1k93373
edited yesterday
answered yesterday
greedoid
26.1k93373
answered yesterday
greedoid
26.1k93373
answered yesterday
greedoid
26.1k93373
26.1k93373
add a comment |Â
add a comment |Â
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set 1, 2, and 3 as local maximum or minimum.
– user295959
yesterday
@user295959 thanks for the help!
– Arka Seth
yesterday