What can be the maximum value of 5-x/x²-10x + 26 for all real x ?

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}


Find the maximum value of the function f(x) = (5-x)/(x^2 - 10x + 26) by simplifying the expression, analyzing critical points, and applying the quotient rule for differentiation. Explore the detailed solution for maximizing this rational function. Suitable for Grades 11-12.

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