1. **Problem statement:** We analyze the variation of the function given its derivative: $$f'(x)=\frac{-5x^2+2x+7}{(x^2-3x+2)^2}$$ 2. **Important observation:** The denominator is $$(x^2-3x+2)^2$$ which is always positive except at $x=1$ and $x=2$ where the function is undefined. Therefore, the sign of $f'(x)$ depends only on the numerator: $$-5x^2+2x+7$$ 3. **Solve numerator = 0:** Set: $$-5x^2+2x+7=0$$ Multiply both sides by $-1$: $$5x^2-2x-7=0$$ Calculate discriminant: $$\Delta=(-2)^2-4\times5\times(-7)=4+140=144$$ Square root: $$\sqrt{144}=12$$ Find roots: $$x=\frac{2\pm12}{10}$$ So, $$x=-1 \quad , \quad x=1.4$$ These are critical points. 4. **Sign of the derivative numerator:** Leading coefficient is $-5$ (negative), so the quadratic opens downward. - $f'(x)>0$ between roots $(-1,1.4)$ - $f'(x)<0$ outside roots $(-\infty,-1)$ and $(1.4,+\infty)$ 5. **Considering domain restrictions:** Function undefined at $x=1$ and $x=2$, so split intervals accordingly. 6. **Final variations:** - On $(-\infty,-1)$: $f'(x)<0$ decreasing - On $(-1,1)$: $f'(x)>0$ increasing - On $(1,1.4)$: $f'(x)>0$ increasing - On $(1.4,2)$: $f'(x)<0$ decreasing - On $(2,+\infty)$: $f'(x)<0$ decreasing 7. **Extrema:** - At $x=-1$, derivative changes from negative to positive, so local minimum: $$f(-1)=-1$$ - At $x=1.4$, derivative changes from positive to negative, so local maximum: $$f(1.4)=24$$ 8. **Summary variation table:** $$ \begin{array}{c|ccccc} x & -\infty & -1 & 1 & 1.4 & 2 & +\infty \\ f'(x) & - & 0 & + & \text{undefined} & + & 0 & - & \text{undefined} & - \\ f(x) & \searrow & \min & \nearrow & \text{asymptote} & \nearrow & \max & \searrow & \text{asymptote} & \searrow \\ \end{array} $$ **Final understanding:** - Function decreases to minimum at $(-1,-1)$ - Increases until asymptote at $x=1$ - After $x=1$, function comes from $-\infty$ and increases to maximum at $(1.4,24)$ - Then decreases toward $-\infty$ at $x=2$ - Finally decreases toward 0 after $x=2$ This confirms the correctness of the variation analysis.