Let `f(x)=x^(3)/3-x^(2)/2+x-16`. Find `f^(')(0), f^(')(-1)`.

Answer

To find the derivatives \( f'(0) \) and \( f'(-1) \) for the function \( f(x) = \frac{x^3}{3} - \frac{x^2}{2} + x - 16 \), we will follow these steps:

### Step 1: Differentiate the function \( f(x) \)

The function is given as:

\[ f(x) = \frac{x^3}{3} - \frac{x^2}{2} + x - 16 \]

We will differentiate \( f(x) \) with respect to \( x \). The derivative of \( x^n \) is given by \( nx^{n-1} \).

1. Differentiate \( \frac{x^3}{3} \):
\[ \frac{d}{dx}\left(\frac{x^3}{3}\right) = \frac{3x^2}{3} = x^2 \]

2. Differentiate \( -\frac{x^2}{2} \):
\[ \frac{d}{dx}\left(-\frac{x^2}{2}\right) = -\frac{2x}{2} = -x \]

3. Differentiate \( x \):
\[ \frac{d}{dx}(x) = 1 \]

4. Differentiate \( -16 \):
\[ \frac{d}{dx}(-16) = 0 \]

Now combine these results:

\[ f'(x) = x^2 - x + 1 \]

### Step 2: Evaluate \( f'(0) \)

Now we will substitute \( x = 0 \) into \( f'(x) \):

\[ f'(0) = 0^2 - 0 + 1 = 1 \]

### Step 3: Evaluate \( f'(-1) \)

Next, we will substitute \( x = -1 \) into \( f'(x) \):

\[ f'(-1) = (-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3 \]

### Final Results

Thus, we have:

\[ f'(0) = 1 \]
\[ f'(-1) = 3 \]