Find the absolute maximum and minimum values of the function over the indicated interval, and indicate the $x$-values at which they occur. \[ f(x)=x+\frac{1}{x} ;[1,25] \] The absolute maximum value is $\frac{626}{25}$ at $x=25$. (Use a comma to separate answers as needed.) The absolute minimum value is at $x=$ (Use a comma to separate answers as needed.)

Step 1 ：The function given is \(f(x)=x+\frac{1}{x}\) and we are to find the absolute maximum and minimum values of the function over the interval [1,25].

Step 2 ：The function is continuous and differentiable on the interval [1,25]. Therefore, the absolute maximum and minimum values of the function occur either at the endpoints of the interval or at critical points in the interval.

Step 3 ：To find the critical points, we need to find the derivative of the function and set it equal to zero. The derivative of the function \(f(x)=x+\frac{1}{x}\) is \(f'(x)=1-\frac{1}{x^2}\).

Step 4 ：Setting this equal to zero gives \(x^2=1\), so \(x=\pm1\). However, only \(x=1\) is in the interval [1,25].

Step 5 ：We then evaluate the function at the endpoints of the interval and at the critical point to find the absolute maximum and minimum values.

Step 6 ：At \(x=1\), \(f(x)=1+\frac{1}{1}=2\).

Step 7 ：At \(x=25\), \(f(x)=25+\frac{1}{25}=\frac{626}{25}\).

Step 8 ：Comparing these values, we find that the absolute maximum value is \(\frac{626}{25}\) at \(x=25\) and the absolute minimum value is \(2\) at \(x=1\).

Step 9 ：\(\boxed{\text{Final Answer: The absolute maximum value is }\frac{626}{25}\text{ at }x=25\text{. The absolute minimum value is }2\text{ at }x=1.}\)