Question

for the function $f(x) = 2x^3 - 3x^2 + 5x + 3$, find $f(x)$. then find $f(0)$ and $f(3)$. to find $f(x)$, first find $f(x)$. $f(x) = 6x^2 - 6x + 5$ find $f(x)$. $f(x) = 12$

Explanation:

Step1: Differentiate \( f'(x) \) to get \( f''(x) \)

Given \( f'(x) = 6x^2 - 6x + 5 \), we use the power rule for differentiation, which states that if \( y = ax^n \), then \( y' = nax^{n - 1} \).
For the term \( 6x^2 \): the derivative is \( 2\times6x^{2 - 1}=12x \).
For the term \( -6x \): the derivative is \( -6 \).
For the constant term \( 5 \): the derivative is \( 0 \).
So, \( f''(x)=12x - 6 \). Wait, but the problem says to find \( f'''(x) \). Let's correct that. First, \( f'(x) = 6x^2 - 6x + 5 \). Then, to find \( f''(x) \) (first derivative of \( f'(x) \)):
Derivative of \( 6x^2 \) is \( 12x \), derivative of \( -6x \) is \( -6 \), derivative of \( 5 \) is \( 0 \). So \( f''(x)=12x - 6 \). Then, to find \( f'''(x) \) (derivative of \( f''(x) \)):
Derivative of \( 12x \) is \( 12 \), derivative of \( -6 \) is \( 0 \). So \( f'''(x)=12 \).

Step2: Find \( f'''(0) \)

Since \( f'''(x) = 12 \) (a constant function), substituting \( x = 0 \) gives \( f'''(0)=12 \).

Step3: Find \( f'''(3) \)

Since \( f'''(x) = 12 \) (a constant function), substituting \( x = 3 \) gives \( f'''(3)=12 \).

Answer:

\( f'''(x)=\boxed{12} \), \( f'''(0)=\boxed{12} \), \( f'''(3)=\boxed{12} \)