Use the first derivative test to determine the location of each local extremum and the value of the function at this extremum. $$ f(x)=\frac{x^{3}}{x^{3}+2} $$

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To find the local extrema of the function $$ f(x) = \frac{x^3}{x^3 + 2} $$ using the first derivative test, we will follow these steps:

1. **Find the first derivative $$ f'(x) $$**.
2. **Determine the critical points by setting $$ f'(x) = 0 $$** and finding where $$ f'(x) $$ is undefined.
3. **Use the first derivative test to identify the nature of each critical point** (whether it's a local maximum, local minimum, or neither).
4. **Evaluate the function at the critical points to find the values of the local extrema**.

Let's start with step 1: finding the first derivative $$ f'(x) $$.

### Step 1: Find the first derivative

Using the quotient rule, where $$ u = x^3 $$ and $$ v = x^3 + 2 $$:

$$
f'(x) = \frac{u'v - uv'}{v^2}
$$

Calculating $$ u' $$ and $$ v' $$:

- $$ u' = 3x^2 $$
- $$ v' = 3x^2 $$

Now substituting into the quotient rule:

$$
f'(x) = \frac{(3x^2)(x^3 + 2) - (x^3)(3x^2)}{(x^3 + 2)^2}
$$

This simplifies to:

$$
f'(x) = \frac{3x^2(x^3 + 2 - x^3)}{(x^3 + 2)^2} = \frac{6x^2}{(x^3 + 2)^2}
$$

### Step 2: Determine the critical points

Set $$ f'(x) = 0 $$:

$$
\frac{6x^2}{(x^3 + 2)^2} = 0
$$

The numerator must be zero:

$$
6x^2 = 0 \implies x = 0
$$

Next, we check where $$ f'(x) $$ is undefined. The denominator $$ (x^3 + 2)^2 $$ is never zero since $$ x^3 + 2 > 0 $$ for all real $$ x $$.

Thus, the only critical point is $$ x = 0 $$.

### Step 3: First derivative test

To determine the nature of the critical point, we will test the sign of $$ f'(x) $$ around $$ x = 0 $$:

- For $$ x < 0 $$ (e.g., $$ x = -1 $$):
  $$
  f'(-1) = \frac{6(-1)^2}{((-1)^3 + 2)^2} = \frac{6}{1} > 0
  $$

- For $$ x > 0 $$ (e.g., $$ x = 1 $$):
  $$
  f'(1) = \frac{6(1)^2}{(1^3 + 2)^2} = \frac{6}{9} > 0
  $$

Since $$ f'(x) > 0 $$ on both sides of $$ x = 0 $$, the function is increasing on both sides of the critical point. Therefore, $$ x = 0 $$ is neither a local maximum nor a local minimum.

### Step 4: Evaluate the function at the critical point

Now, we evaluate $$ f(0) $$:

$$
f(0) = \frac{0^3}{0^3 + 2} = \frac{0}{2} = 0
$$

### Conclusion

The function $$ f(x) = \frac{x^3}{x^3 + 2} $$ has a critical point at $$ x = 0 $$, where the function value is $$ f(0) = 0 $$. However, this point is neither a local maximum nor a local minimum, as the function is increasing on both sides of this point.

Thus, there are no local extrema for this function.
There are no local extrema for the function $$ f(x) = \frac{x^3}{x^3 + 2} $$.

Use the first derivative test to determine the location of each local extremum and the value of the function at this extremum. \( f(x)=\frac{x^{3}}{x^{3}+2} \)

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