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To solve the multiple-choice question (MCQ) based on the function \( f(x) = \frac{2x^3}{3} - \frac{x^2}{2} - 10x \), we need to find the relative extrema (relative minima and maxima).

### Step 1: Find the first derivative \( f'(x) \)
The first derivative of the function will help us determine the critical points where the slope of the tangent to the curve is zero.

\[
f'(x) = \frac{d}{dx} \left( \frac{2x^3}{3} - \frac{x^2}{2} - 10x \right)
\]

Calculating this:

\[
f'(x) = 2x^2 - x - 10
\]

### Step 2: Find the critical points
To find the critical points, set \( f'(x) = 0 \):

\[
2x^2 - x - 10 = 0
\]

This is a quadratic equation. To solve for \( x \), we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 2 \), \( b = -1 \), and \( c = -10 \):

\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-10)}}{2(2)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 80}}{4}
\]
\[
x = \frac{1 \pm \sqrt{81}}{4}
\]
\[
x = \frac{1 \pm 9}{4}
\]

This gives us two critical points:

\[
x = \frac{10}{4} = 2.5 \quad \text{and} \quad x = \frac{-8}{4} = -2
\]

### Step 3: Determine the nature of the critical points
To determine whether each critical point is a relative minimum or maximum, we need to calculate the second derivative \( f''(x) \):

\[
f''(x) = \frac{d}{dx} \left( 2x^2 - x - 10 \right) = 4x - 1
\]

Evaluate \( f''(x) \) at the critical points \( x = 2.5 \) and \( x = -2 \):

- At \( x = 2.5 \):

\[
f''(2.5) = 4(2.5) - 1 = 10 - 1 = 9 > 0
\]
Since \( f''(2.5) > 0 \), \( x = 2.5 \) is a relative minimum.

- At \( x = -2 \):

\[
f''(-2) = 4(-2) - 1 = -8 - 1 = -9 < 0
\]
Since \( f''(-2) < 0 \), \( x = -2 \) is a relative maximum.

### Step 4: Calculate the function values at the critical points
Now, find \( f(2.5) \) and \( f(-2) \):

\[
f(2.5) = \frac{2(2.5)^3}{3} - \frac{(2.5)^2}{2} - 10(2.5)
\]
\[
= \frac{2(15.625)}{3} - \frac{6.25}{2} - 25
\]
\[
= \frac{31.25}{3} - 3.125 - 25 = 10.4167 - 28.125 = -17.7083
\]

\[
f(-2) = \frac{2(-2)^3}{3} - \frac{(-2)^2}{2} - 10(-2)
\]
\[
= \frac{2(-8)}{3} - \frac{4}{2} + 20 = -\frac{16}{3} - 2 + 20
\]
\[
= -5.3333 - 2 + 20 = 12.6667
\]

### Conclusion:
The function \( f(x) \) has a relative minimum at \( x = 2.5 \) and a relative maximum at \( x = -2 \).

**Correct Answer: (c)** relative minimum \( \left( \frac{5}{2}, \frac{-77}{24} \right) \), relative maximum \( \left( -2, \frac{38}{3} \right) \).

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Do you need any further details or have any questions?

Here are some follow-up questions that might be useful:

1. How do you find the critical points of a function?
2. What is the significance of the second derivative in determining relative extrema?
3. Can a critical point be neither a maximum nor a minimum? Explain.
4. How would you find inflection points for the given function?
5. What are the differences between absolute and relative extrema?

**Tip:** Always check the sign of the second derivative at the critical points to determine if they are relative maxima, minima, or points of inflection.

Solve the multiple-choice question (MCQ) based on the function f(x) = 2x^3/3 - x^2/2 - 10x. Determine the relative minima and maxima.

Learn how to solve a calculus problem involving the function f(x) = 2x^3/3 - x^2/2 - 10x by finding its relative minima and maxima using the first and second derivatives. This problem is ideal for students in Grades 11-12.

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