The image presents an application of derivatives to determine intervals where a function is increasing or decreasing, based on the sign of the derivative. The specific example given involves two functions:

1. $f(x) = 2x^3 + 3x^2 - 12x - 7$
2. $f(x) = \frac{x^2}{x - 2}$

Let's solve part (a) first.

Steps to find intervals of increase and decrease for $f(x) = 2x^3 + 3x^2 - 12x - 7$:

1. Find the first derivative $f'(x)$:

   = 6x^2 + 6x - 12$$

2. Solve for critical points by setting $f'(x) = 0$: $6x^2 + 6x - 12 = 0$ Simplifying: $x^2 + x - 2 = 0$ Factoring: $(x - 1)(x + 2) = 0$ So, $x = 1$ and $x = -2$.

3. Determine the sign of $f'(x)$ in intervals divided by the critical points $x = -2$ and $x = 1$:

   • For $x < -2$, choose a test point like $x = -3$: $f'(-3) = 6(-3)^2 + 6(-3) - 12 = 54 - 18 - 12 = 24 \quad (f'(x) > 0)$
   • For $-2 < x < 1$, choose a test point like $x = 0$: $f'(0) = 6(0)^2 + 6(0) - 12 = -12 \quad (f'(x) < 0)$
   • For $x > 1$, choose a test point like $x = 2$: $f'(2) = 6(2)^2 + 6(2) - 12 = 24 + 12 - 12 = 24 \quad (f'(x) > 0)$

4. Conclusion:

   • $f(x)$ is increasing on $(-\infty, -2) \cup (1, \infty)$
   • $f(x)$ is decreasing on $(-2, 1)$

Would you like me to proceed with part (b) or explain any step in more detail?

Here are 5 related questions to deepen your understanding:

1. How does the behavior of the derivative change the shape of the graph?
2. What is the significance of critical points in determining local maxima or minima?
3. Can we conclude whether the critical points are maxima or minima from the derivative alone?
4. How can the second derivative test complement this analysis for concavity?
5. What happens if the derivative does not change sign at a critical point?

Tip: Remember that the derivative test only tells us about local behavior (increase or decrease), but to fully analyze a function, it's helpful to also consider limits and concavity (using the second derivative).