Consider the function $$ f(x) $$ whose second derivative is $$ f^{\prime \prime}(x)=6 x+4 \sin (x) $$. If $$ f(0)=4 $$ and $$ f^{\prime}(0)=2 $$, what is $$ f(5) ? \square $$

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To find $$ f(5) $$, we start with the second derivative of the function:

$$
f''(x) = 6x + 4\sin(x)
$$

We will first integrate $$ f''(x) $$ to find $$ f'(x) $$.

$$
f'(x) = \int f''(x) \, dx = \int (6x + 4\sin(x)) \, dx
$$

We can integrate each term separately:

1. The integral of $$ 6x $$ is:
   $$
   \int 6x \, dx = 3x^2
   $$

2. The integral of $$ 4\sin(x) $$ is:
   $$
   \int 4\sin(x) \, dx = -4\cos(x)
   $$

Combining these results, we have:

$$
f'(x) = 3x^2 - 4\cos(x) + C
$$

where $$ C $$ is a constant of integration. We can find $$ C $$ using the initial condition $$ f'(0) = 2 $$:

$$
f'(0) = 3(0)^2 - 4\cos(0) + C = -4 + C
$$
Setting this equal to 2 gives:

$$
-4 + C = 2 \implies C = 6
$$

Thus, we have:

$$
f'(x) = 3x^2 - 4\cos(x) + 6
$$

Next, we integrate $$ f'(x) $$ to find $$ f(x) $$:

$$
f(x) = \int f'(x) \, dx = \int (3x^2 - 4\cos(x) + 6) \, dx
$$

Again, we integrate each term separately:

1. The integral of $$ 3x^2 $$ is:
   $$
   \int 3x^2 \, dx = x^3
   $$

2. The integral of $$ -4\cos(x) $$ is:
   $$
   \int -4\cos(x) \, dx = -4\sin(x)
   $$

3. The integral of $$ 6 $$ is:
   $$
   \int 6 \, dx = 6x
   $$

Combining these results, we have:

$$
f(x) = x^3 - 4\sin(x) + 6x + D
$$

where $$ D $$ is another constant of integration. We can find $$ D $$ using the initial condition $$ f(0) = 4 $$:

$$
f(0) = (0)^3 - 4\sin(0) + 6(0) + D = D
$$
Setting this equal to 4 gives:

$$
D = 4
$$

Thus, we have:

$$
f(x) = x^3 - 4\sin(x) + 6x + 4
$$

Now, we need to find $$ f(5) $$:

$$
f(5) = 5^3 - 4\sin(5) + 6(5) + 4
$$

Calculating each term:

1. $$ 5^3 = 125 $$
2. $$ 6(5) = 30 $$
3. $$ D = 4 $$

Putting it all together:

$$
f(5) = 125 - 4\sin(5) + 30 + 4 = 159 - 4\sin(5)
$$

Thus, the final answer is:

$$
\boxed{159 - 4\sin(5)}
$$
$$ f(5) = 159 - 4\sin(5) $$

Consider the function \( f(x) \) whose second derivative is \( f^{\prime \prime}(x)=6 x+4 \sin (x) \). If \( f(0)=4 \) and \( f^{\prime}(0)=2 \), what is \( f(5) ? \square \)

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