Respuesta :
Problem: consider the function f(x) whose second derivative is f' '(x)=4x+4sin(x). If f(0)=3 and f'(0)=4, what is f(5)?.
Solution:
Let the function f(x) whose second derivative is:
[tex]f^{\prime\prime}(x)\text{ = 4x+4sin(x)}[/tex]Now, the antiderivative (integral) of the above function would be:
EQUATION 1:
[tex]f^{\prime}(x)=\int f^{\prime\prime}(x)\text{ }dx\text{= }2x^2-4\cos (x)\text{ +C1}[/tex]where C1 is a constant because we have an indefinite integral. Now the antiderivative (integral) of the above function f´(x) is:
[tex]f(x)=\int f^{\prime}(x)\text{ }dx\text{=}\int \text{ (}2x^2-4\cos (x)\text{ +C1)}dx\text{ }[/tex]that is:
EQUATION 2:
[tex]f(x)=\text{ }\frac{2x^3}{3}-4\sin (x)+C1x+\text{ C2}[/tex]where C2 is a constant because we have an indefinite integral.
Now using the previous equation, if f(0)= 3 then:
[tex]3=\text{ C2}[/tex]Now, using equation 1 and the fact that f ´(0) = 4, then we have:
[tex]4=f^{\prime}(0)\text{= }^{}-4\text{ +C1}[/tex]That is:
[tex]4=\text{ }^{}-4\text{ +C1}[/tex]Solve for C1:
[tex]8=\text{ }^{}\text{C1}[/tex]Now, replacing the constants C1 and C2 in equation 2, we have an expression for f(x):
[tex]f(x)=\text{ }\frac{2x^3}{3}-4\sin (x)+8x+3[/tex]Then f(5) would be:
[tex]f(5)=\text{ }\frac{2(5)^3}{3}-4\sin (5)+40+3=\text{ }125.98[/tex]
then the correct answer is:
[tex]f(5)=\text{ }125.98[/tex]