Let's analyze each function one by one using the first derivative test to determine the intervals of increase or decrease, and any local extrema.

Part A: $f(x) = \frac{3x^2 - x - 2}{x - 2}$

The first derivative of the function is given as:

$f'(x) = \frac{3x^2 - 12x + 4}{(x - 2)^2}$

Step 1: Critical Points

To find the critical points, we need to set $f'(x) = 0$. The critical points occur when the numerator equals zero, as the denominator does not contribute to the zero points:

$3x^2 - 12x + 4 = 0$

Now, solve this quadratic equation using the quadratic formula:

$x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(4)}}{2(3)}$ $x = \frac{12 \pm \sqrt{144 - 48}}{6}$ $x = \frac{12 \pm \sqrt{96}}{6}$ $x = \frac{12 \pm 4\sqrt{6}}{6}$ $x = 2 \pm \frac{2\sqrt{6}}{3}$

So, the two critical points are:

$x_1 = 2 + \frac{2\sqrt{6}}{3}, \quad x_2 = 2 - \frac{2\sqrt{6}}{3}$

Step 2: Interval Testing

To determine the intervals of increase and decrease, we will test the sign of $f'(x)$ in intervals around the critical points and the vertical asymptote $x = 2$ (since the denominator $(x - 2)^2$ will approach infinity at $x = 2$, which indicates a vertical asymptote).

Check the sign of $f'(x)$ in the intervals:

• $(-\infty, x_2)$
• $(x_2, 2)$
• $(2, x_1)$
• $(x_1, \infty)$

By plugging test values into $f'(x)$, you can determine if the function is increasing or decreasing in each interval.

Step 3: Local Extrema

The local extrema occur at the critical points. Use the first derivative test:

• If $f'(x)$ changes from positive to negative at a critical point, it is a local maximum.
• If $f'(x)$ changes from negative to positive, it is a local minimum.

Perform these checks for both $x_1$ and $x_2$.

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Part B: $f(x) = x \ln(x^3)$

Simplify the function first:

$f(x) = x \cdot 3\ln(x) = 3x \ln(x)$

The first derivative is given as:

$f'(x) = 3 \ln(x) + 3$

Step 1: Critical Points

Set the derivative equal to zero to find the critical points:

$3 \ln(x) + 3 = 0$ $\ln(x) = -1$ $x = e^{-1} = \frac{1}{e}$

So the critical point is $x = \frac{1}{e}$.

Step 2: Interval Testing

Now, test the sign of $f'(x) = 3 \ln(x) + 3$ around the critical point:

• For $x < \frac{1}{e}$, $\ln(x)$ is negative, so $f'(x) < 0$ (decreasing).
• For $x > \frac{1}{e}$, $\ln(x)$ is positive, so $f'(x) > 0$ (increasing).

Step 3: Local Extrema

At $x = \frac{1}{e}$, $f'(x)$ changes from negative to positive, so it is a local minimum.

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Summary:

• For Part A: The intervals of increase and decrease depend on the signs of the first derivative and the vertical asymptote at $x = 2$. Critical points are $x = 2 \pm \frac{2\sqrt{6}}{3}$, and local extrema need to be confirmed by the first derivative test.
• For Part B: The function is decreasing on $(0, \frac{1}{e})$ and increasing on $(\frac{1}{e}, \infty)$, with a local minimum at $x = \frac{1}{e}$.

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Let me know if you need more details or if you have any questions!

Here are 5 related questions:

1. How do you find vertical asymptotes of rational functions?
2. How can we classify critical points using the second derivative test?
3. What are the conditions for local minima and maxima using the first derivative test?
4. How does the logarithmic function affect the shape of graphs?
5. Can you explain why the derivative is zero at critical points?

Tip: Always simplify the function before differentiating, as it makes the process easier and less error-prone.