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 $\left( -2, \frac{38}{3} \right)$.

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.