To find the function [tex]\( f(x) \)[/tex] given the second derivative [tex]\( f''(x) = 4 - 6x - 40x^3 \)[/tex] and the initial conditions [tex]\( f(0) = 4 \)[/tex] and [tex]\( f'(0) = 3 \)[/tex], we will follow these steps:
1. Integrate the second derivative to find the first derivative:
Given [tex]\( f''(x) = 4 - 6x - 40x^3 \)[/tex], we integrate with respect to [tex]\( x \)[/tex]:
[tex]\[
f'(x) = \int (4 - 6x - 40x^3) \, dx
\][/tex]
This integral is solved as follows:
[tex]\[
f'(x) = \int 4 \, dx - \int 6x \, dx - \int 40x^3 \, dx
\][/tex]
[tex]\[
f'(x) = 4x - 3x^2 - 10x^4 + C_1
\][/tex]
where [tex]\( C_1 \)[/tex] is the constant of integration.
2. Integrate the first derivative to find the original function:
Next, we integrate [tex]\( f'(x) = 4x - 3x^2 - 10x^4 + C_1 \)[/tex]:
[tex]\[
f(x) = \int (4x - 3x^2 - 10x^4 + C_1) \, dx
\][/tex]
This integral is solved as follows:
[tex]\[
f(x) = \int 4x \, dx - \int 3x^2 \, dx - \int 10x^4 \, dx + \int C_1 \, dx
\][/tex]
[tex]\[
f(x) = 2x^2 - x^3 - 2x^5 + C_1x + C_2
\][/tex]
where [tex]\( C_2 \)[/tex] is another constant of integration.
3. Apply the initial conditions to find [tex]\( C_1 \)[/tex] and [tex]\( C_2 \)[/tex]:
Using the initial condition [tex]\( f(0) = 4 \)[/tex]:
[tex]\[
f(0) = 2(0)^2 - (0)^3 - 2(0)^5 + C_1(0) + C_2 = 4
\][/tex]
[tex]\[
C_2 = 4
\][/tex]
Using the initial condition [tex]\( f'(0) = 3 \)[/tex]:
[tex]\[
f'(0) = 4(0) - 3(0)^2 - 10(0)^4 + C_1 = 3
\][/tex]
[tex]\[
C_1 = 3
\][/tex]
4. Substitute the constants back into the function:
Substitute [tex]\( C_1 = 3 \)[/tex] and [tex]\( C_2 = 4 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(x) = 2x^2 - x^3 - 2x^5 + 3x + 4
\][/tex]
Therefore, the original function [tex]\( f(x) \)[/tex] is:
[tex]\[
f(x) = -2x^5 - x^3 + 2x^2 + 3x + 4
\][/tex]
