Consider the function $f\left(x\right)$ whose second derivative is $f^{″}\left(x\right)=6x$. If $f\left(0\right)=2$ and $f^{\prime }\left(0\right)=3$, what is $f\left(x\right)$?

• A. $f\left(x\right)=6x^2+3x+2$
• B. $f\left(x\right)=2x^3+3x+2$
• C. $f\left(x\right)=x^3+3x+2$
• D. $f\left(x\right)=3x^2+2x+2$

Answer: (C)

Verified step by step guidance

1. Step 1: Start by integrating the second derivative f''(x) = 6x to find the first derivative f'(x). The integral of 6x with respect to x is \( \int 6x \, dx = 3x^2 + C_1 \), where \( C_1 \) is the constant of integration.
2. Step 2: Use the given condition f'(0) = 3 to solve for \( C_1 \). Substitute x = 0 into \( f'(x) = 3x^2 + C_1 \), which gives \( f'(0) = C_1 = 3 \). Thus, \( f'(x) = 3x^2 + 3 \).
3. Step 3: Integrate the first derivative f'(x) = 3x^2 + 3 to find the original function f(x). The integral of \( 3x^2 \) is \( x^3 \), and the integral of \( 3 \) is \( 3x \). Adding a constant of integration \( C_2 \), we get \( f(x) = x^3 + 3x + C_2 \).
4. Step 4: Use the given condition f(0) = 2 to solve for \( C_2 \). Substitute x = 0 into \( f(x) = x^3 + 3x + C_2 \), which gives \( f(0) = C_2 = 2 \). Thus, \( f(x) = x^3 + 3x + 2 \).
5. Step 5: Verify the solution by checking that the second derivative of \( f(x) = x^3 + 3x + 2 \) is indeed \( f''(x) = 6x \), and that the initial conditions f(0) = 2 and f'(0) = 3 are satisfied.