1. The problem asks to find the sign of the function $f$ given its derivative $f'(x) = \frac{x^2 - 4x + 3}{(x-2)^2}$. 2. To determine where $f$ is increasing or decreasing, we analyze the sign of $f'(x)$. 3. The denominator $(x-2)^2$ is always positive except at $x=2$ where it is undefined. 4. Focus on the numerator $x^2 - 4x + 3$; factor it: $$x^2 - 4x + 3 = (x-1)(x-3)$$ 5. The critical points from the numerator are $x=1$ and $x=3$. 6. The derivative $f'(x)$ is undefined at $x=2$ due to the denominator. 7. Test intervals determined by points $1$, $2$, and $3$: - For $x < 1$, pick $x=0$: numerator $(0-1)(0-3) = (-1)(-3) = 3 > 0$, denominator positive, so $f'(x) > 0$. - For $1 < x < 2$, pick $x=1.5$: numerator $(1.5-1)(1.5-3) = (0.5)(-1.5) = -0.75 < 0$, denominator positive, so $f'(x) < 0$. - For $2 < x < 3$, pick $x=2.5$: numerator $(2.5-1)(2.5-3) = (1.5)(-0.5) = -0.75 < 0$, denominator positive, so $f'(x) < 0$. - For $x > 3$, pick $x=4$: numerator $(4-1)(4-3) = (3)(1) = 3 > 0$, denominator positive, so $f'(x) > 0$. 8. Summary of $f'(x)$ sign: - Increasing on $(-\infty, 1)$ - Decreasing on $(1, 2)$ and $(2, 3)$ - Increasing on $(3, \infty)$ 9. Note that $f'(x)$ is undefined at $x=2$, so $f$ may have a vertical tangent or discontinuity there. Final answer: $f$ is increasing on $(-\infty, 1)$ and $(3, \infty)$, decreasing on $(1, 2)$ and $(2, 3)$.