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 $$ x= $$ (Type an integer or simplified fraction. Use a comma to separate answers as needed.) 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. (Type an integer or simplified fraction. Use a comma to separate answers as needed.) B. There is no local minimum. What are the inflection points? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The inflection points are at $$ x=\square $$. (Type an integer or simplified fraction. Use a comma to separate answers as needed.)

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**Step 1. Find the derivatives.** We have $$ f(x)=18x(x-1)^3. $$ Differentiate using the product rule: $$ f'(x)=18\,(x-1)^3+18x\cdot3(x-1)^2=18(x-1)^2\Bigl[(x-1)+3x\Bigr]. $$ This simplifies to $$ f'(x)=18(x-1)^2(4x-1). $$ **Step 2. Find the critical points.** Set $$ f'(x)=0 $$: $$ 18(x-1)^2(4x-1)=0. $$ Since $$18\neq0$$, we solve $$ (x-1)^2=0\quad \text{or}\quad 4x-1=0. $$ Thus, the critical points are: $$ x=1\quad \text{and}\quad x=\frac{1}{4}. $$ **Step 3. Determine the nature of the critical points.** Notice that $$(x-1)^2$$ is always nonnegative. Therefore, the sign of $$f'(x)$$ depends entirely on the factor $$4x-1$$. - For $$x < \frac{1}{4}$$, $$4x-1<0$$ so $$f'(x)<0$$. - For $$x>\frac{1}{4}$$ (and $$x\neq1$$), $$4x-1>0$$ so $$f'(x)>0$$. At $$x=\frac{1}{4}$$, $$f'(x)$$ changes from negative to positive. Hence, there is a local minimum at $$x=\frac{1}{4}$$. For $$x = 1$$, note that $$(x-1)^2=0$$ makes $$f'(x)=0$$, but since around $$x=1$$ the factor $$4x-1$$ is positive both immediately to the left and to the right, the derivative does not change sign. Therefore, $$x=1$$ is not a local extremum. **Step 4. Find the second derivative to locate inflection points.** Differentiate $$f'(x)=18(x-1)^2(4x-1)$$ using the product rule. Let $$ u(x)=(x-1)^2 \quad \text{and} \quad v(x)=4x-1. $$ Then $$ u'(x)=2(x-1) \quad \text{and} \quad v'(x)=4. $$ So, $$ f''(x)=18\bigl[u'(x)v(x)+u(x)v'(x)\bigr]=18\left[2(x-1)(4x-1)+4(x-1)^2\right]. $$ Factor out $$2(x-1)$$: $$ f''(x)=18\cdot2(x-1)\Bigl[(4x-1)+2(x-1)\Bigr]=36(x-1)\Bigl[4x-1+2x-2\Bigr]. $$ Simplify the bracket: $$ 4x-1+2x-2=6x-3=3(2x-1). $$ Thus, $$ f''(x)=36(x-1)\cdot3(2x-1)=108(x-1)(2x-1). $$ Set $$ f''(x)=0 $$: $$ 108(x-1)(2x-1)=0. $$ This gives the candidates for inflection points: $$ x-1=0\quad \Rightarrow\quad x=1, $$ $$ 2x-1=0\quad \Rightarrow\quad x=\frac{1}{2}. $$ One would check that the concavity changes at these values, confirming they are inflection points. **Step 5. Summarize the answers.** - **Local maximum:** There is no local maximum. - **Local minimum:** There is a local minimum at $$ x=\frac{1}{4} $$. - **Inflection points:** The inflection points are at $$ x=\frac{1}{2} $$ and $$ x=1 $$. **Summary:** - **Local Maximum:** None - **Local Minimum:** At $$ x = \frac{1}{4} $$ - **Inflection Points:** At $$ x = \frac{1}{2} $$ and $$ x = 1 $$

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