The derivative has a higher grade than the function itself. How is that possible?
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I can't solve a question of a test of a pre-university mathematics course. I understand the rules of derivatives but I am blocked when trying to solve the following question below.
I tried to solve the question by making the derivative of the function and equaling that to the derivative in the question. So, I got that $a= frac21x^4$ etc. But, I may not use those answers as solution for the question.
What am I seeing wrong or where am I blind? Could someone give me a start to solve this question?
This is the question:
If
$$
f(x)=7x^3-2x^2+4x-11
$$ then de derivative
$$ D(f(x))=ax^4+bx^3+cx^2+dx+e$$ What are the values for $a,b,c,d$ and $e$?
Thanks in advance.
derivatives
edited Sep 4 at 6:31
Matti P.
1,497213
asked Sep 4 at 5:40
Casma
61
add a comment |Â
up vote
1
down vote
favorite
I can't solve a question of a test of a pre-university mathematics course. I understand the rules of derivatives but I am blocked when trying to solve the following question below.
I tried to solve the question by making the derivative of the function and equaling that to the derivative in the question. So, I got that $a= frac21x^4$ etc. But, I may not use those answers as solution for the question.
What am I seeing wrong or where am I blind? Could someone give me a start to solve this question?
This is the question:
If
$$
f(x)=7x^3-2x^2+4x-11
$$ then de derivative
$$ D(f(x))=ax^4+bx^3+cx^2+dx+e$$ What are the values for $a,b,c,d$ and $e$?
Thanks in advance.
derivatives
edited Sep 4 at 6:31
Matti P.
1,497213
asked Sep 4 at 5:40
Casma
61
4
Is there any restriction on $a,b,c,d,e$? Why not $a=b=0$? Also, I think $a=frac21x^4$ would not be correct because one typically assumes they are constants, while you are allowing it to be a function of $x$.
– GoodDeeds
Sep 4 at 5:42
4
From the differentiation you should get $$ f'(x) = 21x^2 - 4 x +4 = 0x^4 + 0x^3 + 21x^2 -4x +4 $$
– Matti P.
Sep 4 at 5:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I can't solve a question of a test of a pre-university mathematics course. I understand the rules of derivatives but I am blocked when trying to solve the following question below.
I tried to solve the question by making the derivative of the function and equaling that to the derivative in the question. So, I got that $a= frac21x^4$ etc. But, I may not use those answers as solution for the question.
What am I seeing wrong or where am I blind? Could someone give me a start to solve this question?
This is the question:
If
$$
f(x)=7x^3-2x^2+4x-11
$$ then de derivative
$$ D(f(x))=ax^4+bx^3+cx^2+dx+e$$ What are the values for $a,b,c,d$ and $e$?
Thanks in advance.
derivatives
edited Sep 4 at 6:31
Matti P.
1,497213
asked Sep 4 at 5:40
Casma
61
I can't solve a question of a test of a pre-university mathematics course. I understand the rules of derivatives but I am blocked when trying to solve the following question below.
I tried to solve the question by making the derivative of the function and equaling that to the derivative in the question. So, I got that $a= frac21x^4$ etc. But, I may not use those answers as solution for the question.
What am I seeing wrong or where am I blind? Could someone give me a start to solve this question?
This is the question:
If
$$
f(x)=7x^3-2x^2+4x-11
$$ then de derivative
$$ D(f(x))=ax^4+bx^3+cx^2+dx+e$$ What are the values for $a,b,c,d$ and $e$?
Thanks in advance.
derivatives
derivatives
edited Sep 4 at 6:31
Matti P.
1,497213
asked Sep 4 at 5:40
Casma
61
edited Sep 4 at 6:31
Matti P.
1,497213
asked Sep 4 at 5:40
Casma
61
edited Sep 4 at 6:31
Matti P.
1,497213
edited Sep 4 at 6:31
Matti P.
1,497213
edited Sep 4 at 6:31
Matti P.
1,497213
1,497213
asked Sep 4 at 5:40
Casma
61
asked Sep 4 at 5:40
Casma
61
asked Sep 4 at 5:40
Casma
61
61
4
Is there any restriction on $a,b,c,d,e$? Why not $a=b=0$? Also, I think $a=frac21x^4$ would not be correct because one typically assumes they are constants, while you are allowing it to be a function of $x$.
– GoodDeeds
Sep 4 at 5:42
4
From the differentiation you should get $$ f'(x) = 21x^2 - 4 x +4 = 0x^4 + 0x^3 + 21x^2 -4x +4 $$
– Matti P.
Sep 4 at 5:45
add a comment |Â
4
Is there any restriction on $a,b,c,d,e$? Why not $a=b=0$? Also, I think $a=frac21x^4$ would not be correct because one typically assumes they are constants, while you are allowing it to be a function of $x$.
– GoodDeeds
Sep 4 at 5:42
4
From the differentiation you should get $$ f'(x) = 21x^2 - 4 x +4 = 0x^4 + 0x^3 + 21x^2 -4x +4 $$
– Matti P.
Sep 4 at 5:45
4
4
Is there any restriction on $a,b,c,d,e$? Why not $a=b=0$? Also, I think $a=frac21x^4$ would not be correct because one typically assumes they are constants, while you are allowing it to be a function of $x$.
– GoodDeeds
Sep 4 at 5:42
Is there any restriction on $a,b,c,d,e$? Why not $a=b=0$? Also, I think $a=frac21x^4$ would not be correct because one typically assumes they are constants, while you are allowing it to be a function of $x$.
– GoodDeeds
Sep 4 at 5:42
4
4
From the differentiation you should get $$ f'(x) = 21x^2 - 4 x +4 = 0x^4 + 0x^3 + 21x^2 -4x +4 $$
– Matti P.
Sep 4 at 5:45
From the differentiation you should get $$ f'(x) = 21x^2 - 4 x +4 = 0x^4 + 0x^3 + 21x^2 -4x +4 $$
– Matti P.
Sep 4 at 5:45
add a comment |Â
1 Answer
1
active
oldest
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up vote
1
down vote
The coefficients $a,b,c,d,e$ must be constants. They cannot be dependent on $x$. I think you've gotten hung up on the fact there seems to be a quartic term $(ax^4)$, which you feel shouldn't be there. You can make these terms disappear with zero-coefficients. By ordinary differentiation, you get that
$$D(7x^3-2x^2+4x-11) = 21x^2-4x+4 = underbrace0_ax^4 + underbrace0_bx^3 + underbrace21_cx^2 + underbrace(-4)_dx + underbrace4_e $$
Therefore, the values are $a = 0$, $b = 0$, $c = 21$, $d = -4$, and $e = 4$.
answered Sep 4 at 7:04
Eff
11.1k21537
Thanks a lot and you are right.
– Casma
Sep 4 at 19:55
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The coefficients $a,b,c,d,e$ must be constants. They cannot be dependent on $x$. I think you've gotten hung up on the fact there seems to be a quartic term $(ax^4)$, which you feel shouldn't be there. You can make these terms disappear with zero-coefficients. By ordinary differentiation, you get that
$$D(7x^3-2x^2+4x-11) = 21x^2-4x+4 = underbrace0_ax^4 + underbrace0_bx^3 + underbrace21_cx^2 + underbrace(-4)_dx + underbrace4_e $$
Therefore, the values are $a = 0$, $b = 0$, $c = 21$, $d = -4$, and $e = 4$.
answered Sep 4 at 7:04
Eff
11.1k21537
Thanks a lot and you are right.
– Casma
Sep 4 at 19:55
add a comment |Â
up vote
1
down vote
The coefficients $a,b,c,d,e$ must be constants. They cannot be dependent on $x$. I think you've gotten hung up on the fact there seems to be a quartic term $(ax^4)$, which you feel shouldn't be there. You can make these terms disappear with zero-coefficients. By ordinary differentiation, you get that
$$D(7x^3-2x^2+4x-11) = 21x^2-4x+4 = underbrace0_ax^4 + underbrace0_bx^3 + underbrace21_cx^2 + underbrace(-4)_dx + underbrace4_e $$
Therefore, the values are $a = 0$, $b = 0$, $c = 21$, $d = -4$, and $e = 4$.
answered Sep 4 at 7:04
Eff
11.1k21537
Thanks a lot and you are right.
– Casma
Sep 4 at 19:55
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The coefficients $a,b,c,d,e$ must be constants. They cannot be dependent on $x$. I think you've gotten hung up on the fact there seems to be a quartic term $(ax^4)$, which you feel shouldn't be there. You can make these terms disappear with zero-coefficients. By ordinary differentiation, you get that
$$D(7x^3-2x^2+4x-11) = 21x^2-4x+4 = underbrace0_ax^4 + underbrace0_bx^3 + underbrace21_cx^2 + underbrace(-4)_dx + underbrace4_e $$
Therefore, the values are $a = 0$, $b = 0$, $c = 21$, $d = -4$, and $e = 4$.
answered Sep 4 at 7:04
Eff
11.1k21537
The coefficients $a,b,c,d,e$ must be constants. They cannot be dependent on $x$. I think you've gotten hung up on the fact there seems to be a quartic term $(ax^4)$, which you feel shouldn't be there. You can make these terms disappear with zero-coefficients. By ordinary differentiation, you get that
$$D(7x^3-2x^2+4x-11) = 21x^2-4x+4 = underbrace0_ax^4 + underbrace0_bx^3 + underbrace21_cx^2 + underbrace(-4)_dx + underbrace4_e $$
Therefore, the values are $a = 0$, $b = 0$, $c = 21$, $d = -4$, and $e = 4$.
answered Sep 4 at 7:04
Eff
11.1k21537
answered Sep 4 at 7:04
Eff
11.1k21537
answered Sep 4 at 7:04
Eff
11.1k21537
answered Sep 4 at 7:04
Eff
11.1k21537
11.1k21537
Thanks a lot and you are right.
– Casma
Sep 4 at 19:55
add a comment |Â
Thanks a lot and you are right.
– Casma
Sep 4 at 19:55
Thanks a lot and you are right.
– Casma
Sep 4 at 19:55
Thanks a lot and you are right.
– Casma
Sep 4 at 19:55
add a comment |Â
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4
Is there any restriction on $a,b,c,d,e$? Why not $a=b=0$? Also, I think $a=frac21x^4$ would not be correct because one typically assumes they are constants, while you are allowing it to be a function of $x$.
– GoodDeeds
Sep 4 at 5:42
4
From the differentiation you should get $$ f'(x) = 21x^2 - 4 x +4 = 0x^4 + 0x^3 + 21x^2 -4x +4 $$
– Matti P.
Sep 4 at 5:45