จงหาค่าของ $$ f(x) $$ เมื่อ กำหนด $$ f^{\prime}(x)=6 x^{2}-2 x-5 $$ และ $$ f(0)=2 $$ ก. $$ f(x)=12 x-2 $$ ข. $$ f(x)=6 x-2 $$ ค. $$ f(x)=6 x^{3}-2 x^{2}-5 x+2 $$ ง. $$ f(x)=2 x^{3}-x^{2}-5 x+2 $$ จ. $$ f(x)=2 x^{4}-3 x^{3}+2 x+c $$

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Alright, I need to find the value of $$ f(x) $$ given the derivative $$ f^{\prime}(x) = 6x^{2} - 2x - 5 $$ and the initial condition $$ f(0) = 2 $$. Let's break this down step by step.

First, I recall that to find the original function $$ f(x) $$ from its derivative $$ f^{\prime}(x) $$, I need to perform integration. So, I'll integrate $$ f^{\prime}(x) $$ with respect to $$ x $$.

$$
f(x) = \int (6x^{2} - 2x - 5) \, dx
$$

Integrating term by term:

1. The integral of $$ 6x^{2} $$ is $$ 2x^{3} $$.
2. The integral of $$ -2x $$ is $$ -x^{2} $$.
3. The integral of $$ -5 $$ is $$ -5x $$.

So, combining these, I get:

$$
f(x) = 2x^{3} - x^{2} - 5x + C
$$

Here, $$ C $$ is the constant of integration. To find the value of $$ C $$, I'll use the initial condition $$ f(0) = 2 $$.

Plugging in $$ x = 0 $$:

$$
f(0) = 2(0)^{3} - (0)^{2} - 5(0) + C = 0 - 0 - 0 + C = C
$$

Given that $$ f(0) = 2 $$, it follows that:

$$
C = 2
$$

Therefore, the function $$ f(x) $$ is:

$$
f(x) = 2x^{3} - x^{2} - 5x + 2
$$

Now, let's compare this with the given options:

- 甲. $$ f(x) = 12x - 2 $$
- 乙. $$ f(x) = 6x - 2 $$
- 丙. $$ f(x) = 6x^{3} - 2x^{2} - 5x + 2 $$
- 丁. $$ f(x) = 2x^{3} - x^{2} - 5x + 2 $$
- 戊. $$ f(x) = 2x^{4} - 3x^{3} + 2x + c $$

Looking at the options, 丁. $$ f(x) = 2x^{3} - x^{2} - 5x + 2 $$ matches the function I derived. Therefore, the correct answer is 丁.

丁. $$ f(x) = 2x^{3} - x^{2} - 5x + 2 $$

จงหาค่าของ \( f(x) \) เมื่อ กำหนด \( f^{\prime}(x)=6 x^{2}-2 x-5 \) และ \( f(0)=2 \) ก. \( f(x)=12 x-2 \) ข. \( f(x)=6 x-2 \) ค. \( f(x)=6 x^{3}-2 x^{2}-5 x+2 \) ง. \( f(x)=2 x^{3}-x^{2}-5 x+2 \) จ. \( f(x)=2 x^{4}-3 x^{3}+2 x+c \)

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