The function $f(x)=3+2 x+18 x^{-1}$ has one local minimum and one local maximum. This function has a local maximum at $x=$ with value and a local minimum at $x=$ with value

Step 1 ：Given the function \(f(x) = 3 + 2x + \frac{18}{x}\)

Step 2 ：Find the derivative of the function: \(f'(x) = 2 - \frac{18}{x^2}\)

Step 3 ：Set the derivative equal to zero and solve for x to find the critical points: \(x = -3, 3\)

Step 4 ：Find the second derivative of the function: \(f''(x) = \frac{36}{x^3}\)

Step 5 ：Substitute the critical points into the second derivative. If the result is positive, the function has a local minimum at that x-value. If the result is negative, the function has a local maximum at that x-value.

Step 6 ：Substituting \(x = -3\) into the second derivative gives a negative result, indicating a local maximum at \(x = -3\)

Step 7 ：Substituting \(x = 3\) into the second derivative gives a positive result, indicating a local minimum at \(x = 3\)

Step 8 ：Substitute the critical points into the original function to find the corresponding y-values: \(f(-3) = -9\), \(f(3) = 15\)

Step 9 ：\(\boxed{\text{Final Answer: The function has a local maximum at } x = -3 \text{ with value } -9 \text{ and a local minimum at } x = 3 \text{ with value } 15}\)