Question

Let f (x) be a polynomial function such that f (x) + f'(x) + f''(x) = x5 + 64. Find the polynomial f(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f

Answer 1

x^5 - 5x^4 + 60x^2 - 120x + 64

Answer 2

To find the polynomial function $f(x)$, we are given the equation $f(x) + f'(x) + f''(x) = x^5 + 64$. The general form of the polynomial $f(x)$ is provided as $f(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f$.

1. Find the derivatives of $f(x)$: Given $f(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f$.

The first derivative, $f'(x)$, is: $f'(x) = \frac{d}{dx}(x^5 + bx^4 + cx^3 + dx^2 + ex + f)$ $f'(x) = 5x^4 + 4bx^3 + 3cx^2 + 2dx + e$

The second derivative, $f''(x)$, is: $f''(x) = \frac{d}{dx}(5x^4 + 4bx^3 + 3cx^2 + 2dx + e)$ $f''(x) = 20x^3 + 12bx^2 + 6cx + 2d$

2. Substitute $f(x)$, $f'(x)$, and $f''(x)$ into the given equation: $f(x) + f'(x) + f''(x) = x^5 + 64$ $(x^5 + bx^4 + cx^3 + dx^2 + ex + f) + (5x^4 + 4bx^3 + 3cx^2 + 2dx + e) + (20x^3 + 12bx^2 + 6cx + 2d) = x^5 + 64$

3. Group terms by powers of $x$ and equate coefficients: $x^5$: $1 = 1$ (This confirms the leading coefficient)

$x^4$: $b + 5 = 0 \implies b = -5$

$x^3$: $c + 4b + 20 = 0$ Substitute $b = -5$: $c + 4(-5) + 20 = 0$ $c - 20 + 20 = 0 \implies c = 0$

$x^2$: $d + 3c + 12b = 0$ Substitute $b = -5$ and $c = 0$: $d + 3(0) + 12(-5) = 0$ $d + 0 - 60 = 0 \implies d = 60$

$x^1$: $e + 2d + 6c = 0$ Substitute $c = 0$ and $d = 60$: $e + 2(60) + 6(0) = 0$ $e + 120 + 0 = 0 \implies e = -120$

Constant term: $f + e + 2d = 64$ Substitute $d = 60$ and $e = -120$: $f + (-120) + 2(60) = 64$ $f - 120 + 120 = 64 \implies f = 64$

4. Construct the polynomial $f(x)$ using the found coefficients: Substitute the values $b = -5$, $c = 0$, $d = 60$, $e = -120$, and $f = 64$ into the general form of $f(x)$: $f(x) = x^5 + (-5)x^4 + (0)x^3 + (60)x^2 + (-120)x + (64)$ $f(x) = x^5 - 5x^4 + 60x^2 - 120x + 64$

The final answer is $\boxed{x^5 - 5x^4 + 60x^2 - 120x + 64}$.

Explanation of the solution:

1. Assume $f(x)$ is a polynomial of degree 5, as suggested by the RHS $x^5+64$ and the nature of the equation $f(x)+f'(x)+f''(x)$. The given form $f(x) = x^5 + bx^4 + cx^3 + dx^2 + ex + f$ aligns with this.
2. Calculate the first and second derivatives, $f'(x)$ and $f''(x)$, respectively.
3. Substitute $f(x)$, $f'(x)$, and $f''(x)$ into the given differential equation $f(x) + f'(x) + f''(x) = x^5 + 64$.
4. Collect terms by powers of $x$ on the left side of the equation.
5. Equate the coefficients of corresponding powers of $x$ on both sides of the equation. This generates a system of linear equations for the unknown coefficients $b, c, d, e, f$.
6. Solve this system of equations to find the values of $b, c, d, e, f$.
7. Substitute these values back into the assumed polynomial form of $f(x)$ to obtain the final answer.