To determine which statement is true, we need to analyze the function $f'(x) = x^3 - 3x - 2$. This is the first derivative of the function $f(x)$, and we are interested in finding the critical points and their nature (whether they represent relative minima, maxima, or points of inflection).

Step 1: Find the critical points

Critical points occur where the first derivative $f'(x)$ is equal to 0 or undefined. To find the critical points, solve the equation:

$f'(x) = 0 \quad \Rightarrow \quad x^3 - 3x - 2 = 0$

Let's factor or solve this cubic equation.

Step 2: Solve $x^3 - 3x - 2 = 0$

By trial and error or using the Rational Root Theorem, we can check possible rational roots (which are factors of -2 and 1). Let's start with $x = -1$:

$(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$

So, $x = -1$ is a root. We can factor out $(x + 1)$ from the cubic polynomial using synthetic division or long division.

Performing synthetic division on $x^3 - 3x - 2$ by $x + 1$, we get:

$x^3 - 3x - 2 = (x + 1)(x^2 - x - 2)$

Now, factor the quadratic $x^2 - x - 2$:

$x^2 - x - 2 = (x - 2)(x + 1)$

So, the factored form of the cubic equation is:

$x^3 - 3x - 2 = (x + 1)^2(x - 2)$

Step 3: Critical points

The critical points are where the first derivative equals 0, i.e., the values of $x$ that satisfy $(x + 1)^2(x - 2) = 0$.

Thus, the critical points are:

• $x = -1$ (with multiplicity 2)
• $x = 2$

Step 4: Determine the nature of the critical points using the second derivative

We will now compute the second derivative of $f(x)$ to determine the concavity at each critical point.

Given: $f'(x) = x^3 - 3x - 2$

We differentiate again to find $f''(x)$: $f''(x) = 3x^2 - 3$

Now, evaluate $f''(x)$ at the critical points.

• At $x = -1$: $f''(-1) = 3(-1)^2 - 3 = 3 - 3 = 0$ This result suggests that we cannot conclude directly whether $x = -1$ is a relative minimum, maximum, or point of inflection using the second derivative test, so we must examine the first derivative's sign around this point.

• At $x = 2$: $f''(2) = 3(2)^2 - 3 = 12 - 3 = 9$ Since $f''(2) > 0$, the point $x = 2$ is a relative minimum.

Step 5: Analyze the behavior around $x = -1$

Since $x = -1$ has a multiplicity of 2 (i.e., $(x + 1)^2$ in the factorization), the function doesn't change sign around $x = -1$. Thus, $x = -1$ is a point of inflection, not a relative minimum or maximum.

Conclusion

• There is one relative minimum at $x = 2$.
• There is no relative maximum at $x = -1$ because it is an inflection point.

Thus, the correct answer is: f has one relative minimum and no relative maxima.

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Would you like to dive deeper into any of these steps, or do you have more questions on this?

Here are a few related questions:

1. What is the second derivative test and how is it used to classify critical points?
2. How do you factor cubic polynomials?
3. How do multiplicity of roots affect the behavior of the graph of a function?
4. How can we determine whether a critical point is a minimum or maximum using the first derivative test?
5. What is the role of the Rational Root Theorem in solving cubic equations?

Tip: The second derivative test is often more reliable when you have a simple critical point. For points where the second derivative is 0, you may need to examine the first derivative’s sign around the point or use higher derivatives.