Find the points of local maxima and local minima, if any, of the following functions. Find also the local maximum and local minimum values :
`g(x)=(1)/(x^(2)+2)`

Answer

To find the points of local maxima and minima for the function \( g(x) = \frac{1}{x^2 + 2} \), we will follow these steps:

### Step 1: Find the first derivative \( g'(x) \)

Using the quotient rule, where \( u = 1 \) and \( v = x^2 + 2 \):
\[ g'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]
Here, \( u' = 0 \) and \( v' = 2x \).

Substituting these into the formula gives:
\[ g'(x) = \frac{(x^2 + 2)(0) - (1)(2x)}{(x^2 + 2)^2} = \frac{-2x}{(x^2 + 2)^2} \]

### Step 2: Set the first derivative equal to zero to find critical points

Setting \( g'(x) = 0 \):
\[ \frac{-2x}{(x^2 + 2)^2} = 0 \]
This implies:
\[ -2x = 0 \implies x = 0 \]

### Step 3: Determine the second derivative \( g''(x) \)

Now we will find the second derivative to determine the nature of the critical point. We differentiate \( g'(x) \):
\[ g'(x) = \frac{-2x}{(x^2 + 2)^2} \]
Using the quotient rule again:
Let \( u = -2x \) and \( v = (x^2 + 2)^2 \).

Then,
\[ g''(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]
Where \( u' = -2 \) and \( v' = 2(x^2 + 2)(2x) = 4x(x^2 + 2) \).

Substituting these into the formula gives:
\[ g''(x) = \frac{(x^2 + 2)^2(-2) - (-2x)(4x(x^2 + 2))}{(x^2 + 2)^4} \]
Simplifying this:
\[ g''(x) = \frac{-2(x^2 + 2)^2 + 8x^2(x^2 + 2)}{(x^2 + 2)^4} \]
\[ = \frac{-2(x^2 + 2) + 8x^2}{(x^2 + 2)^3} \]
\[ = \frac{6x^2 - 4}{(x^2 + 2)^3} \]

### Step 4: Evaluate the second derivative at the critical point \( x = 0 \)

Now substituting \( x = 0 \):
\[ g''(0) = \frac{6(0)^2 - 4}{(0^2 + 2)^3} = \frac{-4}{2^3} = \frac{-4}{8} = -\frac{1}{2} \]

Since \( g''(0) < 0 \), this indicates that there is a local maximum at \( x = 0 \).

### Step 5: Find the maximum value of the function

Now we find the value of the function at this critical point:
\[ g(0) = \frac{1}{0^2 + 2} = \frac{1}{2} \]

### Conclusion

Thus, the function \( g(x) = \frac{1}{x^2 + 2} \) has:
- A local maximum at \( x = 0 \) with a maximum value of \( g(0) = \frac{1}{2} \).
- There are no local minima since the function is always positive and approaches 0 as \( x \) approaches \( \pm \infty \).