Find the absolute extrema of the function on the closed interval. $$ g(x)=\frac{2 x^{2}}{x-6},[-3,2] $$ minimum (smaller $$ x $$-value) $$ \quad(x, y)= $$ ( $$ \square $$ ) minimum (larger $$ x $$-value) $$ \quad(x, y)= $$ $$ \square $$ 1) maximum $$ (x, y)=( $$ $$ \square $$ )

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We are given
$$
g(x)=\frac{2x^{2}}{x-6}
$$
on the interval $$[-3,2]$$. We first note that the denominator $$x-6$$ is nonzero on $$[-3,2]$$ (since $$6$$ is not in the interval), so the function is continuous on the closed interval and the Extreme Value Theorem applies.

### Step 1. Find the Critical Points

Differentiate $$g(x)$$ using the quotient rule. Write
$$
g(x)=\frac{2x^2}{x-6}.
$$
The quotient rule gives:
$$
g'(x)=\frac{(2x^2)'(x-6)-2x^2\,(x-6)'}{(x-6)^2}.
$$
Calculate the derivatives:
$$
(2x^2)'=4x \quad \text{and} \quad (x-6)'=1.
$$
Thus,
$$
g'(x)=\frac{4x(x-6)-2x^2}{(x-6)^2}.
$$
Simplify the numerator:
$$
4x(x-6)-2x^2 = 4x^2-24x-2x^2 = 2x^2-24x.
$$
Factor out $$2x$$:
$$
2x^2-24x=2x(x-12).
$$
Therefore,
$$
g'(x)=\frac{2x(x-12)}{(x-6)^2}.
$$

Set the numerator equal to zero:
$$
2x(x-12)=0.
$$
Thus, the possible critical points are:
$$
x=0 \quad \text{or} \quad x=12.
$$
Since $$12$$ is not in the interval $$[-3,2]$$, the only critical point in the interval is $$x=0$$.

### Step 2. Evaluate the Function at Endpoints and Critical Points

- **At $$x = -3$$:**
  $$
  g(-3)=\frac{2(-3)^2}{-3-6}=\frac{18}{-9}=-2.
  $$

- **At $$x = 0$$:**
  $$
  g(0)=\frac{2\cdot 0^2}{0-6}=0.
  $$

- **At $$x = 2$$:**
  $$
  g(2)=\frac{2(2)^2}{2-6}=\frac{8}{-4}=-2.
  $$

### Step 3. Determine the Absolute Extrema

- The **absolute maximum** value is $$0$$ at $$x=0$$. Hence,
  $$
  (x,y)=(0,0).
  $$

- The **absolute minimum** value is $$-2$$ which occurs at both endpoints. Since the problem distinguishes between:
  - Minimum at the smaller $$x$$-value: at $$x=-3$$,
    $$
    (x,y)=(-3,-2).
    $$
  - Minimum at the larger $$x$$-value: at $$x=2$$,
    $$
    (x,y)=(2,-2).
    $$

### Final Answers

- Minimum (smaller $$x$$-value):
  $$
  (-3,-2)
  $$
  
- Minimum (larger $$x$$-value):
  $$
  (2,-2)
  $$
  
- Maximum:
  $$
  (0,0)
  $$
The absolute maximum of $$ g(x) = \frac{2x^2}{x-6} $$ on $$[-3,2]$$ is $$0$$ at $$x=0$$. The absolute minimum is $$-2$$ at both $$x=-3$$ and $$x=2$$. - Minimum (smaller $$x$$-value): $$(-3, -2)$$ - Minimum (larger $$x$$-value): $$(2, -2)$$ - Maximum: $$(0, 0)$$

Find the absolute extrema of the function on the closed interval.

minimum (smaller -value) ( )
minimum (larger -value) 1)
maximum

)

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Find the absolute extrema of the function on the closed interval. minimum (smaller -value) ( ) minimum (larger -value) 1) maximum )