1. **State the problem:** We have a function defined as $f(x) = x^3 - x^2 - 8x + 5$ for $x < a$, and we know $f$ is increasing on this domain. We need to find the largest possible value of $a$. 2. **Recall the rule for increasing functions:** A function is increasing on an interval if its derivative is non-negative on that interval. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x^3 - x^2 - 8x + 5) = 3x^2 - 2x - 8$$ 4. **Set the derivative to be non-negative for $x < a$:** $$3x^2 - 2x - 8 \geq 0$$ 5. **Solve the quadratic inequality:** First find roots of $3x^2 - 2x - 8 = 0$ using the quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 96}}{6} = \frac{2 \pm 10}{6}$$ So roots are: $$x_1 = \frac{2 - 10}{6} = -\frac{8}{6} = -\frac{4}{3}$$ $$x_2 = \frac{2 + 10}{6} = 2$$ 6. **Analyze the sign of $f'(x)$:** Since the leading coefficient $3$ is positive, the parabola opens upward. - For $x < -\frac{4}{3}$, $f'(x) > 0$ - For $-\frac{4}{3} < x < 2$, $f'(x) < 0$ - For $x > 2$, $f'(x) > 0$ 7. **Determine the largest $a$ such that $f$ is increasing on $(-\infty, a)$:** To have $f'(x) \geq 0$ for all $x < a$, the interval $(-\infty, a)$ must lie entirely where $f'(x) \geq 0$. From the sign analysis, $f'(x) \geq 0$ for $x < -\frac{4}{3}$ and for $x > 2$. Since the domain is $x < a$, to keep $f$ increasing on $(-\infty, a)$, $a$ must be at most the right endpoint of the first positive interval, which is $a = -\frac{4}{3}$. If $a$ were larger than $-\frac{4}{3}$, the function would include values where $f'(x) < 0$, violating the increasing condition. **Final answer:** $$a = -\frac{4}{3}$$