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. Factor and Identify Zeros**

We are given
$$
f(x)=18x(x-1)^3.
$$
Setting $$ f(x)=0 $$,
$$
18x(x-1)^3=0,
$$
which yields the zeros:
$$
x=0 \quad \text{and} \quad x=1.
$$

**Step 2. End Behavior**

Expanding the function shows the leading term comes from
$$
18x\cdot(x-1)^3 = 18x^4 - \cdots.
$$
Since the highest power is $$ x^4 $$ with a positive coefficient, the function behaves as follows:
- As $$ x\to \infty $$, $$ f(x)\to +\infty $$.
- As $$ x\to -\infty $$, $$ f(x)\to +\infty $$.

**Step 3. Find the Critical Numbers**

Differentiate $$ f(x) $$ using the product rule:
$$
f(x)=18x(x-1)^3.
$$
Differentiate:
$$
f'(x)=18\Big[(x-1)^3+ x\cdot3(x-1)^2\Big]=18\Big[(x-1)^2\big((x-1)+3x\big)\Big].
$$
Simplify:
$$
(x-1)+3x=4x-1,
$$
so that
$$
f'(x)=18(x-1)^2(4x-1).
$$

Setting $$ f'(x)=0 $$ gives:
$$
(x-1)^2(4x-1)=0.
$$
Thus, the critical numbers are:
- $$ x=1 $$ (a double root from $$(x-1)^2$$),
- $$ x=\frac{1}{4} $$ (from $$4x-1=0$$).

**Step 4. Classify the Critical Numbers**

Examine the sign changes of $$ f'(x) $$:

- **At $$ x=\frac{1}{4} $$:**  
  Choose a test value slightly less than $$ \frac{1}{4} $$, say $$ x=0 $$:  
  $$
  f'(0)=18(0-1)^2(4\cdot0-1)=18(1)(-1)=-18\quad (\text{negative}).
  $$
  Choose a test value slightly greater than $$ \frac{1}{4} $$, say $$ x=0.3 $$:  
  $$
  (0.3-1)^2>0 \quad \text{and} \quad 4(0.3)-1=1.2-1=0.2>0,
  $$
  so $$ f'(0.3)>0 $$.  
  Since $$ f'(x) $$ changes from negative to positive at $$ x=\frac{1}{4} $$, we have a **local minimum** at $$ x=\frac{1}{4} $$.

- **At $$ x=1 $$:**  
  The factor $$ (x-1)^2 $$ is squared, so near $$ x=1 $$ the sign of $$ f'(x) $$ is determined only by $$ (4x-1) $$, which is positive around $$ x=1 $$ (since $$ 4(1)-1=3>0 $$).  
  Testing values: for $$ x=0.9 $$ and $$ x=1.1 $$, $$ f'(x) $$ remains positive, meaning there is no sign change.  
  Hence, there is no local extremum at $$ x=1 $$.

**Step 5. Evaluate the Local Minimum Value**

To find the minimum function value at $$ x=\frac{1}{4} $$:
$$
f\left(\frac{1}{4}\right)=18\left(\frac{1}{4}\right)\left(\frac{1}{4}-1\right)^3.
$$
Calculate:
$$
\frac{1}{4}-1=-\frac{3}{4}, \quad \text{so} \quad \left(-\frac{3}{4}\right)^3=-\frac{27}{64}.
$$
Thus,
$$
f\left(\frac{1}{4}\right)=18\cdot\frac{1}{4}\cdot\left(-\frac{27}{64}\right)=-\frac{18\cdot27}{4\cdot64}=-\frac{486}{256}=-\frac{243}{128}.
$$

**Step 6. Graph Summary**

- **Intercepts:** $$ f(x)=0 $$ at $$ x=0 $$ and $$ x=1 $$.
- **End Behavior:**  
  $$ f(x)\to +\infty $$ as $$ x\to \pm\infty $$.
- **Critical Points:**  
  A local minimum at $$ x=\frac{1}{4} $$ with $$ f\left(\frac{1}{4}\right)=-\frac{243}{128} $$.  
  No local maximum.
- **Other Features:**  
  The factor $$ (x-1)^3 $$ indicates that the graph has a point of inflection or a change in concavity at $$ x=1 $$ even though it is also an $$ x $$-intercept.

**Final Answers**

- Local Maximum:  
  B. There is no local maximum.

- Local Minimum:  
  A. The local minimum is at $$ x=\frac{1}{4} $$.
There is no local maximum. The local minimum is at $$ x=\frac{1}{4} $$.

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