To prove and determine [tex]\( k \)[/tex] for the given function [tex]\( f(x) = \frac{15 + x^2}{x - 1} \)[/tex], we need to use the quotient rule for differentiation and simplify the result to match the given form.
### Step-by-Step Solution:
#### 1. Define the function and apply the quotient rule:
The function is:
[tex]\[ f(x) = \frac{15 + x^2}{x - 1} \][/tex]
The quotient rule states:
[tex]\[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{[v(x)]^2} \][/tex]
For [tex]\( f(x) = \frac{u(x)}{v(x)} \)[/tex]:
- [tex]\( u(x) = 15 + x^2 \)[/tex]
- [tex]\( v(x) = x - 1 \)[/tex]
#### 2. Compute derivatives [tex]\( u'(x) \)[/tex] and [tex]\( v'(x) \)[/tex]:
[tex]\[ u'(x) = \frac{d}{dx}(15 + x^2) = 2x \][/tex]
[tex]\[ v'(x) = \frac{d}{dx}(x - 1) = 1 \][/tex]
#### 3. Apply the quotient rule:
[tex]\[
f'(x) = \frac{(2x)(x - 1) - (15 + x^2)(1)}{(x - 1)^2}
\][/tex]
Simplify the numerator:
[tex]\[
(2x)(x - 1) - (15 + x^2) = 2x^2 - 2x - 15 - x^2 = x^2 - 2x - 15
\][/tex]
Thus,
[tex]\[
f'(x) = \frac{x^2 - 2x - 15}{(x - 1)^2}
\][/tex]
#### 4. Simplify and compare to the given form:
We need to show that this equals:
[tex]\[
\frac{(x-5)(x+k)}{x^2 - 2x + 1}
\][/tex]
Notice that [tex]\( (x - 1)^2 = x^2 - 2x + 1 \)[/tex], so the denominator already matches.
Now, we need to factor the numerator [tex]\( x^2 - 2x - 15 \)[/tex] and determine [tex]\( k \)[/tex]:
[tex]\[
x^2 - 2x - 15 = (x - 5)(x + 3)
\][/tex]
Thus:
[tex]\[
f'(x) = \frac{(x - 5)(x + 3)}{x^2 - 2x + 1}
\][/tex]
By comparing this with [tex]\( \frac{(x-5)(x+k)}{x^2 -2x + 1} \)[/tex], we can see that:
[tex]\[
x + k = x + 3 \implies k = 3
\][/tex]
### Conclusion:
Using the quotient rule and comparing the simplified forms, we determined that the value of [tex]\( k \)[/tex] is [tex]\( \boxed{3} \)[/tex].
