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$P(x)$ is a polynomial of least degree with a local maximum of $6$ at $x=1$ and minimum of $2$ at $x=3$. Find $P^prime(0)$.
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Let $P(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and minimum at $x=3$.If $P(1)=6$ and $P(3)=2$, then find $P'(0)$.
Method: Since the smallest polynomial whose derivative gives 2 root would be a cubic equation. So I assumed my function to be
$$P(x) = ax^3+ bx^2+cx+d$$
Solving the condition above gives me $b=-6a$ and $c=11a$. I am stuck after this. Can anyone tell me how to proceed from here?
calculus
edited 7 mins ago
Blue
44.4k868142
asked 26 mins ago
Abhinav
1325
add a comment |Â
up vote
2
down vote
favorite
Let $P(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and minimum at $x=3$.If $P(1)=6$ and $P(3)=2$, then find $P'(0)$.
Method: Since the smallest polynomial whose derivative gives 2 root would be a cubic equation. So I assumed my function to be
$$P(x) = ax^3+ bx^2+cx+d$$
Solving the condition above gives me $b=-6a$ and $c=11a$. I am stuck after this. Can anyone tell me how to proceed from here?
calculus
edited 7 mins ago
Blue
44.4k868142
asked 26 mins ago
Abhinav
1325
2
Well, you also know that $P(1)=6,P(3)=2$.
– lulu
25 mins ago
1
Note: I don't agree that $c=11a$. You have $P'(x)=3ax^2+2bx+c=lambda(x-3)(x-1)=lambda x^2-4lambda x+3lambdaimplies a=3lambda,,2b=-4lambda=-12a,,c=3lambda=9a$. No?
– lulu
21 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $P(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and minimum at $x=3$.If $P(1)=6$ and $P(3)=2$, then find $P'(0)$.
Method: Since the smallest polynomial whose derivative gives 2 root would be a cubic equation. So I assumed my function to be
$$P(x) = ax^3+ bx^2+cx+d$$
Solving the condition above gives me $b=-6a$ and $c=11a$. I am stuck after this. Can anyone tell me how to proceed from here?
calculus
edited 7 mins ago
Blue
44.4k868142
asked 26 mins ago
Abhinav
1325
Let $P(x)$ be a real polynomial of least degree which has a local maximum at $x=1$ and minimum at $x=3$.If $P(1)=6$ and $P(3)=2$, then find $P'(0)$.
Method: Since the smallest polynomial whose derivative gives 2 root would be a cubic equation. So I assumed my function to be
$$P(x) = ax^3+ bx^2+cx+d$$
Solving the condition above gives me $b=-6a$ and $c=11a$. I am stuck after this. Can anyone tell me how to proceed from here?
calculus
calculus
edited 7 mins ago
Blue
44.4k868142
asked 26 mins ago
Abhinav
1325
edited 7 mins ago
Blue
44.4k868142
asked 26 mins ago
Abhinav
1325
edited 7 mins ago
Blue
44.4k868142
edited 7 mins ago
Blue
44.4k868142
edited 7 mins ago
Blue
44.4k868142
44.4k868142
asked 26 mins ago
Abhinav
1325
asked 26 mins ago
Abhinav
1325
asked 26 mins ago
Abhinav
1325
1325
2
Well, you also know that $P(1)=6,P(3)=2$.
– lulu
25 mins ago
1
Note: I don't agree that $c=11a$. You have $P'(x)=3ax^2+2bx+c=lambda(x-3)(x-1)=lambda x^2-4lambda x+3lambdaimplies a=3lambda,,2b=-4lambda=-12a,,c=3lambda=9a$. No?
– lulu
21 mins ago
add a comment |Â
2
Well, you also know that $P(1)=6,P(3)=2$.
– lulu
25 mins ago
1
Note: I don't agree that $c=11a$. You have $P'(x)=3ax^2+2bx+c=lambda(x-3)(x-1)=lambda x^2-4lambda x+3lambdaimplies a=3lambda,,2b=-4lambda=-12a,,c=3lambda=9a$. No?
– lulu
21 mins ago
2
2
Well, you also know that $P(1)=6,P(3)=2$.
– lulu
25 mins ago
Well, you also know that $P(1)=6,P(3)=2$.
– lulu
25 mins ago
1
1
Note: I don't agree that $c=11a$. You have $P'(x)=3ax^2+2bx+c=lambda(x-3)(x-1)=lambda x^2-4lambda x+3lambdaimplies a=3lambda,,2b=-4lambda=-12a,,c=3lambda=9a$. No?
– lulu
21 mins ago
Note: I don't agree that $c=11a$. You have $P'(x)=3ax^2+2bx+c=lambda(x-3)(x-1)=lambda x^2-4lambda x+3lambdaimplies a=3lambda,,2b=-4lambda=-12a,,c=3lambda=9a$. No?
– lulu
21 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Hint: Consider the System
$$a+b+c+d=6$$
$$27a+9b+3c+d=2$$
$$3a+2b+c=0$$
$$27a+6b+c=0$$
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
add a comment |Â
up vote
2
down vote
HINT
You attempt was right to assume a function of the type $P(x)=ax^3+bx^2+cx+d$ as the solution. Rewriting the conditions as $P(1)=6$,$P(3)=2$,$P'(1)=0$ and $P'(3)=0$ leads to a system of four equations with four variables $a,b,c$ and $d$.
Can you do it on your own from hereon?
answered 20 mins ago
mrtaurho
801219
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Hint: Consider the System
$$a+b+c+d=6$$
$$27a+9b+3c+d=2$$
$$3a+2b+c=0$$
$$27a+6b+c=0$$
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
add a comment |Â
up vote
3
down vote
accepted
Hint: Consider the System
$$a+b+c+d=6$$
$$27a+9b+3c+d=2$$
$$3a+2b+c=0$$
$$27a+6b+c=0$$
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Hint: Consider the System
$$a+b+c+d=6$$
$$27a+9b+3c+d=2$$
$$3a+2b+c=0$$
$$27a+6b+c=0$$
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
Hint: Consider the System
$$a+b+c+d=6$$
$$27a+9b+3c+d=2$$
$$3a+2b+c=0$$
$$27a+6b+c=0$$
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
answered 21 mins ago
Dr. Sonnhard Graubner
69.2k32761
69.2k32761
add a comment |Â
add a comment |Â
up vote
2
down vote
HINT
You attempt was right to assume a function of the type $P(x)=ax^3+bx^2+cx+d$ as the solution. Rewriting the conditions as $P(1)=6$,$P(3)=2$,$P'(1)=0$ and $P'(3)=0$ leads to a system of four equations with four variables $a,b,c$ and $d$.
Can you do it on your own from hereon?
answered 20 mins ago
mrtaurho
801219
add a comment |Â
up vote
2
down vote
HINT
You attempt was right to assume a function of the type $P(x)=ax^3+bx^2+cx+d$ as the solution. Rewriting the conditions as $P(1)=6$,$P(3)=2$,$P'(1)=0$ and $P'(3)=0$ leads to a system of four equations with four variables $a,b,c$ and $d$.
Can you do it on your own from hereon?
answered 20 mins ago
mrtaurho
801219
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT
You attempt was right to assume a function of the type $P(x)=ax^3+bx^2+cx+d$ as the solution. Rewriting the conditions as $P(1)=6$,$P(3)=2$,$P'(1)=0$ and $P'(3)=0$ leads to a system of four equations with four variables $a,b,c$ and $d$.
Can you do it on your own from hereon?
answered 20 mins ago
mrtaurho
801219
HINT
You attempt was right to assume a function of the type $P(x)=ax^3+bx^2+cx+d$ as the solution. Rewriting the conditions as $P(1)=6$,$P(3)=2$,$P'(1)=0$ and $P'(3)=0$ leads to a system of four equations with four variables $a,b,c$ and $d$.
Can you do it on your own from hereon?
answered 20 mins ago
mrtaurho
801219
answered 20 mins ago
mrtaurho
801219
answered 20 mins ago
mrtaurho
801219
answered 20 mins ago
mrtaurho
801219
801219
add a comment |Â
add a comment |Â
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2
Well, you also know that $P(1)=6,P(3)=2$.
– lulu
25 mins ago
1
Note: I don't agree that $c=11a$. You have $P'(x)=3ax^2+2bx+c=lambda(x-3)(x-1)=lambda x^2-4lambda x+3lambdaimplies a=3lambda,,2b=-4lambda=-12a,,c=3lambda=9a$. No?
– lulu
21 mins ago