If `f(x)=2x^(2)+3x-5,` then what is `f'(0)+3f'(-1)` equal to?

Options

A. `-1`
B. 0
C. 1
D. 2

Explanation

To solve the problem, we need to find the value of \( f'(0) + 3f'(-1) \) for the function \( f(x) = 2x^2 + 3x - 5 \).

### Step 1: Find the derivative of \( f(x) \).

The function is given as:
\[ f(x) = 2x^2 + 3x - 5 \]

To find the derivative \( f'(x) \), we apply the power rule:
\[ f'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(-5) \]

Calculating each term:
- The derivative of \( 2x^2 \) is \( 4x \).
- The derivative of \( 3x \) is \( 3 \).
- The derivative of a constant \( -5 \) is \( 0 \).

Combining these results, we have:
\[ f'(x) = 4x + 3 \]

### Step 2: Calculate \( f'(0) \).

Now we substitute \( x = 0 \) into the derivative:
\[ f'(0) = 4(0) + 3 = 3 \]

### Step 3: Calculate \( f'(-1) \).

Next, we substitute \( x = -1 \) into the derivative:
\[ f'(-1) = 4(-1) + 3 = -4 + 3 = -1 \]

### Step 4: Compute \( f'(0) + 3f'(-1) \).

Now we can compute:
\[ f'(0) + 3f'(-1) = 3 + 3(-1) = 3 - 3 = 0 \]

### Final Answer:
Thus, the value of \( f'(0) + 3f'(-1) \) is:
\[ \boxed{0} \]