We are asked to find the maximum value of the function $f(x) = \frac{5-x}{x^2 - 10x + 26}$ for all real values of $x$.

Step 1: Simplify the function

The given function is: $f(x) = \frac{5-x}{x^2 - 10x + 26}$

The denominator can be simplified by completing the square:

$x^2 - 10x + 26 = (x^2 - 10x + 25) + 1 = (x - 5)^2 + 1$

Thus, the function becomes: $f(x) = \frac{5-x}{(x-5)^2 + 1}$

Step 2: Consider the maximum possible value of $f(x)$

We analyze the function to find its maximum value. The numerator is $5-x$, and the denominator is always positive since it is the sum of a square and 1.

Step 3: Finding the derivative and setting it to zero

To find the critical points, we differentiate $f(x)$ with respect to $x$:

$f(x) = \frac{5-x}{(x-5)^2 + 1}$

Using the quotient rule:

$f'(x) = \frac{-1 \cdot ((x-5)^2 + 1) - (5-x) \cdot 2(x-5)}{((x-5)^2 + 1)^2}$

Simplifying the numerator:

$f'(x) = \frac{-((x-5)^2 + 1) + 2(5-x)(x-5)}{((x-5)^2 + 1)^2}$

Expanding the terms:

$f'(x) = \frac{-((x-5)^2 + 1) + 2(5-x)(x-5)}{((x-5)^2 + 1)^2}$ [ f'(x) = \frac{-(x^2 - 10x + 25 + 1) + 2(25 - 10x + x^2)}{((x-5)^2 + 1)^2}