Question

Consider the function f(X) whose second derivative is f''(x) = 2x + 4sin(x). If f(0) = 3 and f'(0) = 4, what is f(x)

Answer 1

so I know the anti derivative of the function is \[x^2 -4\cos(x) + c\]

Answer 2

Now we need to find that constant knowing that f'(0) = 4 so we set it up like\[f(0) = 0^2 -4\cos(0) + c = 3\]

Answer 3

scratch that it = 4 so in this case c = 0

Answer 4

f'(x) = x ^2- 4cos(x)+c not f(x)

Answer 5

okay

Answer 6

then I find the derivative of that f(x) is \[\frac{ x^3 }{ 3 } - 4\sin(x) + c\]

Answer 7

\[f^{'}(0)=4\]
\[f^{'}(0)=(0)^2-4cos(0)+c=4\]

Answer 8

find C first and then integrate

Answer 9

c = 0

Answer 10

-4 * 1 = 4

Answer 11

wait

Answer 12

8?

Answer 13

\[c=8\]
\[f^{'}(x)=x^2-4cos(x)+8\]

Answer 14

correct

Answer 15

right right, then we get \[f(x) = \frac{ x^3 }{ 3 } - 4\sin(x) + 8x + c\]

Answer 16

correct

Answer 17

so c = 3 because everything 0s out

Answer 18

yes

Answer 19

thanks for catching that error

Answer 20

np