For the function: $$ f(x)=4 x-6 x^{2 / 3} $$ on the interval $$ [-1,3] $$, The absolute min occurs at $$ ([?],[]) $$ The absolute max occurs at $$ ([],[]) $$

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To find the absolute minimum and maximum of the function $$ f(x) = 4x - 6x^{2/3} $$ on the interval $$ [-1, 3] $$, we will follow these steps:

1. **Find the critical points** by taking the derivative of the function and setting it to zero.
2. **Evaluate the function** at the critical points and the endpoints of the interval.
3. **Compare the values** to determine the absolute minimum and maximum.

### Step 1: Find the derivative

First, we need to find the derivative $$ f'(x) $$:

$$
f'(x) = \frac{d}{dx}(4x - 6x^{2/3})
$$

Let's calculate the derivative.
Simplify the expression by following steps:
- step0: Solution:
  $$4-6\times \frac{2}{3}x^{\frac{-1}{3}}$$
- step1: Rewrite the fraction:
  $$4-6\times \frac{2}{3}x^{-\frac{1}{3}}$$
- step2: Multiply the terms:
  $$4-4x^{-\frac{1}{3}}$$
- step3: Simplify:
  $$4-\frac{4\sqrt[3]{x^{2}}}{x}$$
- step4: Reduce fractions to a common denominator:
  $$\frac{4x}{x}-\frac{4\sqrt[3]{x^{2}}}{x}$$
- step5: Transform the expression:
  $$\frac{4x-4\sqrt[3]{x^{2}}}{x}$$
The derivative of the function is:

$$
f'(x) = 4 - 4x^{-1/3}
$$

### Step 2: Set the derivative to zero

Now, we set the derivative equal to zero to find the critical points:

$$
4 - 4x^{-1/3} = 0
$$

Solving for $$ x $$:

$$
4 = 4x^{-1/3}
$$

Dividing both sides by 4:

$$
1 = x^{-1/3}
$$

Taking the reciprocal:

$$
x^{1/3} = 1
$$

Cubing both sides:

$$
x = 1
$$

### Step 3: Evaluate the function at critical points and endpoints

Now we will evaluate $$ f(x) $$ at the critical point $$ x = 1 $$ and the endpoints $$ x = -1 $$ and $$ x = 3 $$.

1. **At $$ x = -1 $$**:
   $$
   f(-1) = 4(-1) - 6(-1)^{2/3} = -4 - 6(1) = -4 - 6 = -10
   $$

2. **At $$ x = 1 $$**:
   $$
   f(1) = 4(1) - 6(1)^{2/3} = 4 - 6(1) = 4 - 6 = -2
   $$

3. **At $$ x = 3 $$**:
   $$
   f(3) = 4(3) - 6(3)^{2/3} = 12 - 6(3^{2/3})
   $$
   First, we calculate $$ 3^{2/3} $$:
   $$
   3^{2/3} = (3^{1/3})^2 \approx (1.442)^2 \approx 2.08
   $$
   Now substituting back:
   $$
   f(3) \approx 12 - 6(2.08) = 12 - 12.48 \approx -0.48
   $$

### Step 4: Compare the values

Now we have the following values:

- $$ f(-1) = -10 $$
- $$ f(1) = -2 $$
- $$ f(3) \approx -0.48 $$

### Conclusion

- The **absolute minimum** occurs at $$ x = -1 $$ with $$ f(-1) = -10 $$.
- The **absolute maximum** occurs at $$ x = 3 $$ with $$ f(3) \approx -0.48 $$.

Thus, we can summarize:

- The absolute min occurs at $$ ([-1],[-10]) $$
- The absolute max occurs at $$ ([3],[-0.48]) $$
The absolute minimum occurs at $$ (-1, -10) $$ and the absolute maximum occurs at $$ (3, -0.48) $$.

For the function: \( f(x)=4 x-6 x^{2 / 3} \) on the interval \( [-1,3] \), The absolute min occurs at \( ([?],[]) \) The absolute max occurs at \( ([],[]) \)

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