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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1. Write the function: $$ f(x)=18x(x-1)^3 $$ 2. Compute the derivative using the product rule: $$ f'(x)=18(x-1)^3+18x\cdot3(x-1)^2 $$ Factor out the common term $$18(x-1)^2$$: $$ f'(x)=18(x-1)^2\left[(x-1)+3x\right]=18(x-1)^2(4x-1) $$ 3. Find the critical points by setting $$f'(x)=0$$: $$ 18(x-1)^2(4x-1)=0 $$ This gives: - $$ (x-1)^2=0 $$ $$\Rightarrow x=1 $$ - $$ 4x-1=0 $$ $$\Rightarrow x=\frac{1}{4} $$ 4. Determine the nature of each critical point: - For $$ x=\frac{1}{4} $$: Test the sign of $$f'(x)$$: - For $$ x<\frac{1}{4} $$, $$4x-1<0$$ and $$(x-1)^2$$ is positive, so $$f'(x)<0$$. - For $$ x>\frac{1}{4} $$, $$4x-1>0$$, so $$f'(x)>0$$. Since $$f'(x)$$ changes from negative to positive, there is a local minimum at $$ x=\frac{1}{4} $$. - For $$x=1$$: Since $$(x-1)^2$$ is zero at $$x=1$$, $$f'(1)=0$$, however, note that $$(x-1)^2$$ is nonnegative and the sign of $$f'(x)$$ is determined solely by $$4x-1$$ which is positive both just left and right of $$x=1$$. Hence, the function does not change from increasing to decreasing or vice versa, and $$ x=1 $$ is not a local extremum. 5. Summarize the results: - Local maximum: There is no local maximum. - Local minimum: The local minimum is at $$ x=\frac{1}{4} $$. 6. Sketching the graph (pertinent features): - **Zeros and factors:** $$ f(x)=18x(x-1)^3 $$ Zeros at $$ x=0 $$ and $$ x=1 $$ (with $$ x=1 $$ having multiplicity 3, which implies a point of inflection with a horizontal tangent). - **Critical point:** - Local minimum at $$ x=\frac{1}{4} $$ with $$ f\left(\frac{1}{4}\right) $$ computed as: $$ f\left(\frac{1}{4}\right)=18\left(\frac{1}{4}\right)\left(\frac{1}{4}-1\right)^3=18\left(\frac{1}{4}\right)\left(-\frac{3}{4}\right)^3 $$ $$ =\frac{18}{4}\left(-\frac{27}{64}\right)=-\frac{486}{256}=-\frac{243}{128} $$ - **Behavior at $$ x=1 $$:** - Although $$ f'(1)=0 $$, since the function keeps increasing through $$x=1$$, this is not an extremum but a horizontal inflection point. Final Answers: Local maximum: B. There is no local maximum. Local minimum: A. The local minimum is at $$ x=\frac{1}{4} $$. Local maximum: B. There is no local maximum. Local minimum: A. The local minimum is at $$ x=\frac{1}{4} $$.

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