Consider `f'(x)=(x^(2))/(2)-kx+1` such that `f(0)=0` and `f(3)=15`.
`f''(-(2)/(3))` is equal to

Options

A. `-1`
B. `(1)/(3)`
C. `(1)/(2)`
D. `1`

Answer

To solve the problem, we need to find \( f''\left(-\frac{2}{3}\right) \) given that \( f'(x) = \frac{x^2}{2} - kx + 1 \), \( f(0) = 0 \), and \( f(3) = 15 \).

### Step 1: Integrate \( f'(x) \) to find \( f(x) \)

We start by integrating \( f'(x) \):

\[ f(x) = \int f'(x) \, dx = \int \left( \frac{x^2}{2} - kx + 1 \right) \, dx \]

\[ = \frac{x^3}{6} - \frac{kx^2}{2} + x + C \]

### Step 2: Use the condition \( f(0) = 0 \) to find \( C \)

Substituting \( x = 0 \):

\[ f(0) = \frac{0^3}{6} - \frac{k(0^2)}{2} + 0 + C = 0 \]

This implies:

\[ C = 0 \]

Thus, we have:

\[ f(x) = \frac{x^3}{6} - \frac{kx^2}{2} + x \]

### Step 3: Use the condition \( f(3) = 15 \) to find \( k \)

Now we substitute \( x = 3 \):

\[ f(3) = \frac{3^3}{6} - \frac{k(3^2)}{2} + 3 = 15 \]

Calculating \( f(3) \):

\[ f(3) = \frac{27}{6} - \frac{9k}{2} + 3 = 15 \]

Simplifying \( \frac{27}{6} = 4.5 \) and \( 3 = 3 \):

\[ 4.5 - \frac{9k}{2} + 3 = 15 \]

Combining terms:

\[ 7.5 - \frac{9k}{2} = 15 \]

Subtracting 7.5 from both sides:

\[ -\frac{9k}{2} = 15 - 7.5 \]

\[ -\frac{9k}{2} = 7.5 \]

Multiplying both sides by -2:

\[ 9k = -15 \]

Thus:

\[ k = -\frac{15}{9} = -\frac{5}{3} \]

### Step 4: Find \( f''(x) \)

Now we differentiate \( f'(x) \) to find \( f''(x) \):

\[ f'(x) = \frac{x^2}{2} - kx + 1 \]

Differentiating gives:

\[ f''(x) = x - k \]

Substituting \( k = -\frac{5}{3} \):

\[ f''(x) = x + \frac{5}{3} \]

### Step 5: Evaluate \( f''\left(-\frac{2}{3}\right) \)

Now we substitute \( x = -\frac{2}{3} \):

\[ f''\left(-\frac{2}{3}\right) = -\frac{2}{3} + \frac{5}{3} \]

Calculating:

\[ f''\left(-\frac{2}{3}\right) = \frac{-2 + 5}{3} = \frac{3}{3} = 1 \]

### Final Answer

Thus, \( f''\left(-\frac{2}{3}\right) = 1 \).

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