Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ y=f(x) $$. $$ f(x)=18 x(x-1)^{3} $$ What is/are the local maximum/a? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The local maximum/a is/are at $$ x=\square $$ 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. 10 . Use a comma to separate answers as needed.) A. The local minimum/a is/are at $$ x= $$. (Type an integer or simplified fraction. Use a comma to separate answers as needed.) B. There is no local minimum.

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**Step 1. Find the Critical Points** We have $$ f(x)=18x(x-1)^3. $$ Differentiate $$ f(x) $$ using the product rule: $$ f'(x)=18(x-1)^3+18x\cdot 3(x-1)^2. $$ Factor out $$ 18(x-1)^2 $$: $$ f'(x)=18(x-1)^2\left[(x-1)+3x\right]=18(x-1)^2(4x-1). $$ Set $$ f'(x)=0 $$: $$ 18(x-1)^2(4x-1)=0. $$ This yields: $$ (x-1)^2=0 \quad \Rightarrow \quad x=1, $$ and $$ 4x-1=0 \quad \Rightarrow \quad x=\frac{1}{4}. $$ --- **Step 2. Classify the Critical Points** Examine the sign of $$ f'(x) $$: $$ f'(x)=18(x-1)^2(4x-1). $$ - The factor $$ (x-1)^2 $$ is always nonnegative. - The sign of $$ f'(x) $$ is determined by $$ (4x-1) $$. 1. For $$ x<\frac{1}{4} $$: $$ 4x-1<0 \quad \Rightarrow \quad f'(x)<0 \quad (\text{since } (x-1)^2>0). $$ Thus, $$ f(x) $$ is decreasing. 2. For $$\frac{1}{4}0 \quad \Rightarrow \quad f'(x)>0. \] Thus, $$ f(x) $$ is increasing. At $$ x=\frac{1}{4} $$, the derivative changes from negative to positive. Therefore, there is a **local minimum** at $$ x=\frac{1}{4} $$. At $$ x=1 $$, note that even though $$ f'(1)=0 $$, the sign of $$ f'(x) $$ (driven solely by $$ 4x-1 $$) remains positive on both sides of $$ x=1 $$ (since for $$ x $$ just less than 1 and $$ x $$ just greater than 1, $$ 4x-1 $$ is positive). Therefore, there is **no local maximum** or minimum at $$ x=1 $$; this is a point of inflection or a saddle point. --- **Step 3. Answer the Questions** - **Local Maximum:** **Choice B.** There is no local maximum. - **Local Minimum:** **Choice A.** The local minimum is at $$ x=\frac{1}{4}. $$ **Local Maximum:** Choice B. There is no local maximum. **Local Minimum:** Choice A. The local minimum is at $$ x=\frac{1}{4} $$.

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