Find the critical numbers a) $$ f(x)=x^{3}-6 x^{2}+9 x $$ b) $$ f(x)=\frac{x^{2}}{x-2} $$ Type $$ x= $$ lower number,x=higher number Blank \# $$ 1 \square $$ Blank \# $$ 2 \square $$

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**a) $$ f(x)=x^3-6x^2+9x $$**

1. Compute the derivative:
   $$
   f'(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3)
   $$

2. Set the derivative equal to 0:
   $$
   3(x-1)(x-3)=0 \quad \Longrightarrow \quad x-1=0 \quad \text{or} \quad x-3=0
   $$
   Thus, the critical numbers are:
   $$
   x=1 \quad \text{and} \quad x=3
   $$

Since $$1$$ is lower than $$3$$:

- Blank \# $$1 \square$$ $$x=1$$
- Blank \# $$2 \square$$ $$x=3$$

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**b) $$ f(x)=\frac{x^2}{x-2} $$**

1. Note the domain: The function is undefined at $$ x=2 $$ (since division by zero is not allowed), so $$ x\ne2 $$.

2. Compute the derivative using the quotient rule. If
   $$
   u=x^2 \quad \text{and} \quad v=x-2,
   $$
   then
   $$
   u'=2x \quad \text{and} \quad v'=1.
   $$
   The quotient rule gives:
   $$
   f'(x)=\frac{u'v-uv'}{v^2}=\frac{2x(x-2)-x^2\cdot1}{(x-2)^2}=\frac{2x^2-4x-x^2}{(x-2)^2}=\frac{x^2-4x}{(x-2)^2}.
   $$

3. Factor the numerator:
   $$
   x^2-4x=x(x-4)
   $$

4. Set the derivative equal to 0. Since the denominator $$ (x-2)^2 $$ is never 0 for $$ x\ne2 $$, we set:
   $$
   x(x-4)=0 \quad \Longrightarrow \quad x=0 \quad \text{or} \quad x=4.
   $$

Both $$ x=0 $$ and $$ x=4 $$ are in the domain.

Since $$0$$ is lower than $$4$$:

- Blank \# $$1 \square$$ $$x=0$$
- Blank \# $$2 \square$$ $$x=4$$
**a) $$ f(x)=x^3-6x^2+9x $$** - Blank \# $$1 \square$$ $$x=1$$ - Blank \# $$2 \square$$ $$x=3$$ **b) $$ f(x)=\frac{x^2}{x-2} $$** - Blank \# $$1 \square$$ $$x=0$$ - Blank \# $$2 \square$$ $$x=4$$

Find the critical numbers a) \( f(x)=x^{3}-6 x^{2}+9 x \) b) \( f(x)=\frac{x^{2}}{x-2} \) Type \( x= \) lower number,x=higher number Blank \# \( 1 \square \) Blank \# \( 2 \square \)

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