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)=(x)/(5)+(5)/(x),x > 0,`

Answer

To find the points of local maxima and minima for the function \( g(x) = \frac{x}{5} + \frac{5}{x} \) where \( x > 0 \), we will follow these steps:

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

The first step is to differentiate the function \( g(x) \).

\[ g(x) = \frac{x}{5} + \frac{5}{x} \]

Using the power rule and the quotient rule, we find the derivative:

\[ g'(x) = \frac{1}{5} - \frac{5}{x^2} \]

### Step 2: Set the first derivative equal to zero

To find the critical points, we set the first derivative \( g'(x) \) equal to zero:

\[ \frac{1}{5} - \frac{5}{x^2} = 0 \]

### Step 3: Solve for \( x \)

Rearranging the equation gives:

\[ \frac{5}{x^2} = \frac{1}{5} \]

Cross-multiplying leads to:

\[ 5 \cdot 5 = x^2 \implies x^2 = 25 \implies x = 5 \quad (\text{since } x > 0) \]

### Step 4: Find the second derivative \( g''(x) \)

Next, we find the second derivative to determine the nature of the critical point:

\[ g'(x) = \frac{1}{5} - \frac{5}{x^2} \]

Differentiating again:

\[ g''(x) = 0 + \frac{10}{x^3} = \frac{10}{x^3} \]

### Step 5: Evaluate the second derivative at the critical point

Now we evaluate \( g''(x) \) at \( x = 5 \):

\[ g''(5) = \frac{10}{5^3} = \frac{10}{125} = \frac{2}{25} > 0 \]

Since \( g''(5) > 0 \), this indicates that the function has a local minimum at \( x = 5 \).

### Step 6: Find the local minimum value

Finally, we find the value of \( g(x) \) at the critical point \( x = 5 \):

\[ g(5) = \frac{5}{5} + \frac{5}{5} = 1 + 1 = 2 \]

### Conclusion

Thus, the function \( g(x) \) has a local minimum at \( x = 5 \) with a minimum value of \( g(5) = 2 \).

### Summary of Results

- Local minimum point: \( x = 5 \)
- Local minimum value: \( g(5) = 2 \)