Summarize the pertinent information obtaired by applying the graphing strategy and sketch the graph of $$ y=f(x) $$. $$ f(x)=2 x(x-3)^{3} $$ What is/are the local maximum/a? Select the correct choice below and, it necessary, ill in the answer box to complele your choce. A. The local maximuma islare at $$ x= $$ $$ \square $$ $$ \square $$. (Type an inleger or simplifed fraction. Use a comma 10 separate answers as neoded.) B. There is no local maximum. What is/are the local minimum/a? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The local minmurn/a is/are at $$ x= $$ $$ \square $$ (Type an inleger or simplitied fraction. Use a cumma to separate answers as noeded.) B. There is no local minimum. What are the inflection points? Select the correct choice below and, if necossary, filt in the answer box to complele your chcice. A. The inflection points are of $$ x= $$ $$ \square $$ . (Type an inleger or simplified iraction. Use a comma to separate answers as needed)

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**Step 1. Find the derivative** We have $$ f(x)=2x(x-3)^3. $$ Differentiate using the product rule: $$ f'(x)=2(x-3)^3+2x\cdot 3(x-3)^2. $$ Factor out the common term $$2(x-3)^2$$: $$ f'(x)=2(x-3)^2\Big[(x-3)+3x\Big]=2(x-3)^2(4x-3). $$ **Step 2. Find the critical points** Set $$f'(x)=0$$: $$ 2(x-3)^2(4x-3)=0. $$ This gives: 1. $$ (x-3)^2=0 \Rightarrow x=3$$. 2. $$4x-3=0 \Rightarrow x=\frac{3}{4}$$. **Step 3. Determine the nature of each critical point** Since the factor $$(x-3)^2$$ is always nonnegative, the sign of $$f'(x)$$ is determined by $$4x-3$$. - For $$x<\frac{3}{4}$$, we have $$4x-3<0$$ so $$f'(x)<0$$. - For $$x>\frac{3}{4}$$ (at least until $$x=3$$), we have $$4x-3>0$$ so $$f'(x)>0$$. Thus, at $$x=\frac{3}{4}$$ the derivative changes from negative to positive. This indicates a local minimum at $$x=\frac{3}{4}$$. At $$x=3$$, although $$f'(3)=0$$, the factor $$ (x-3)^2$$ is zero but does not change sign (since it is always nonnegative). In fact, for $$x$$ just below and just above 3, the sign of $$f'(x)$$ is positive. Hence, $$x=3$$ is not a local extremum but may be a point of inflection. **Step 4. Find the second derivative (to locate inflection points)** Differentiate $$f'(x)=2(4x-3)(x-3)^2$$ using the product rule: $$ f''(x)=2\left[4\,(x-3)^2+ (4x-3)\cdot 2(x-3)\right]. $$ Simplify: $$ f''(x)=8(x-3)^2+4(4x-3)(x-3). $$ Factor out $$4(x-3)$$: $$ f''(x)=4(x-3)\Big[2(x-3)+(4x-3)\Big]. $$ Combine like terms: $$ 2(x-3)+(4x-3)=2x-6+4x-3=6x-9. $$ Thus, $$ f''(x)=4(x-3)(6x-9)=12(x-3)(2x-3). $$ **Step 5. Find possible inflection points** Inflection points occur when $$f''(x)=0$$: $$ 12(x-3)(2x-3)=0. $$ This gives: 1. $$x-3=0\Rightarrow x=3$$. 2. $$2x-3=0\Rightarrow x=\frac{3}{2}$$. Check the sign changes of $$f''(x)$$ around these points to confirm inflection: - Near $$x=3$$: For $$x<3$$ (e.g. $$x=2.9$$), $$(x-3)<0$$ and $$(2x-3)>0$$ so $$f''(x)<0$$; for $$x>3$$ (e.g. $$x=3.1$$), both factors are positive so $$f''(x)>0$$. Hence, $$x=3$$ is an inflection point. - Near $$x=\frac{3}{2}$$: One can similarly check that the concavity changes. Thus, $$x=\frac{3}{2}$$ is also an inflection point. **Step 6. Summary of Findings** - **Local Maximum:** Since there is no change in derivative sign from positive to negative at any critical point, there is **no local maximum**. - **Local Minimum:** Occurs at $$x=\frac{3}{4}$$. - **Inflection Points:** Occur at $$x=\frac{3}{2}$$ and $$x=3$$. **Final Answers** 1. Local maximum: **B. There is no local maximum.** 2. Local minimum: **A. The local minimum is at $$ x = \frac{3}{4} $$.** 3. Inflection points: **A. The inflection points are at $$ x = \frac{3}{2},\, 3 $$.** **Summary of Findings:** - **Local Maximum:** None - **Local Minimum:** At $$ x = \frac{3}{4} $$ - **Inflection Points:** At $$ x = \frac{3}{2} $$ and $$ x = 3 $$ **Final Answers:** 1. Local maximum: B. There is no local maximum. 2. Local minimum: A. The local minimum is at $$ x = \frac{3}{4} $$. 3. Inflection points: A. The inflection points are at $$ x = \frac{3}{2},\, 3 $$.

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