Since the sum of those must be identically equal to ,
then the largest power of x that occurs in the sum, which is n+1, must be
the largest power that occurs in , which is 4. Therefore,
n+1 = 4 and n=3. Substituting (and eliminating the 3 dots since they
are no longer needed):
Simplifying,
Since the coefficients of x3 and x are 0 in ,
and .
Since the coefficient of x2 is 13,
and since its constant term is 2,
So we have the system of equations
Your teacher must have meant for you to take the constant
of integration to be 0. Otherwise you have 5 unknowns but
only 4 equations. I'll use my system solver to solve that.
I get:
So that makes
Edwin
.
f (x) is polynomial function, f '(x) + int f (x) dx = x ^4 + 13 x ^2 + 2,
Find f(x)
~~~~~~~~~~~~~~~~~~~~~~
We want to find f(x) as a polynomial f(x) = + + . . . + .
Taking derivative decreases the degree of a polynomial by one unit.
Taking antiderivative increases the degree of a polynomial by one unit.
Since the sum f ' (x) + int f (x) dx is a polynomial of degree 4,
----------------------
it means that the sough polynomial f(x) is of degree 3:
f(x) = ax^3 + bx^2 + cx + d.
Then
f ' (x) = + 2bx + c,
int f(x) dx = + + + dx + E.
So, in the sum f ' (x) + int f(x) dx
----------------------
(a) coefficient at is It gives an equation
= 1; hence a = 4.
(b) coefficient at is 0. It gives an equation
= 0; hence b = 0.
(c) coefficient at is 13. It gives an equation
= 13, or = 13 ---> = 13 - 12 = 1 ---> c = 2.
(d) coefficient at is 0. It gives an equation
2b + d = 0, which implies 2*0 + d = 0; hence, d = 0.
+------------------------------------------------------------+
| At this point, the problem is just solved to the end. |
| a = 4; b = 0; c = 2; d = 0. |
+------------------------------------------------------------+
The sough polynomial is f(x) = 4x^3 + 2x. ANSWER
CHECK. The derivative is f ' (x) = .
The anti-derivative is F(x) = = .
The sum f ' (x) + F(x) = + = . ! correct !
Solved.
Do not accept any other answer.
--------------
The solution by Edwin is INCORRECT.
To make sure that it is incorrect, simply take the antiderivative of his leading term .
This antiderivative is = , and no other arguments are needed anymore.