The problem in the image reads:

Exercise 3

Let $f$ be a function that is strictly decreasing on $\mathbb{R}$. If the curve $C_f$ passes through the point $A(6, -2)$, solve the inequality: $f(x^2 - x) < -2.$

Solution

1. Since $f$ is a strictly decreasing function, $f(x) < f(y)$ when $x > y$.

2. The curve $C_f$ passes through the point $A(6, -2)$, which tells us that: $f(6) = -2.$

3. We need to solve the inequality: $f(x^2 - x) < -2.$

   Since $f(6) = -2$, we want $x^2 - x > 6$ to ensure $f(x^2 - x) < f(6) = -2$ due to the decreasing nature of $f$.

4. Solve $x^2 - x > 6$: $x^2 - x - 6 > 0.$

5. Factorize the quadratic inequality: $(x - 3)(x + 2) > 0.$

6. Determine the intervals where this inequality holds:

   • The roots of $(x - 3)(x + 2) = 0$ are $x = 3$ and $x = -2$.
   • Use a sign chart or test intervals around these points:
     • For $x < -2$: $(x - 3)(x + 2) > 0$.
     • For $-2 < x < 3$: $(x - 3)(x + 2) < 0$.
     • For $x > 3$: $(x - 3)(x + 2) > 0$.

   Thus, $(x - 3)(x + 2) > 0$ when $x < -2$ or $x > 3$.

Solution Set

The solution to the inequality $f(x^2 - x) < -2$ is: $x \in (-\infty, -2) \cup (3, \infty).$

Would you like further details on any part of this solution, or have any questions?

Here are some related questions for further understanding:

1. What happens if $f$ is an increasing function instead?
2. How does knowing $f(6) = -2$ affect the solution?
3. How can we interpret $f(x^2 - x) < -2$ graphically?
4. What if the inequality were $f(x^2 - x) > -2$?
5. How can we solve similar inequalities with other types of functions?

Tip: When dealing with inequalities involving monotonic functions, use their properties (increasing or decreasing) to determine the inequality's direction.