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.

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 \( x=\frac{1}{4} \).