Question

find f(x) for the function f(x) = \frac{x + 9}{x^2 - 7x + 2}. f(x) =

Explanation:

Step1: Identify the quotient rule

The function \( f(x) = \frac{x + 9}{x^2 - 7x + 2} \) is a quotient of two functions, \( u(x)=x + 9 \) and \( v(x)=x^2 - 7x + 2 \). The quotient rule states that \( f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \).

Step2: Find \( u'(x) \) and \( v'(x) \)

For \( u(x)=x + 9 \), the derivative \( u'(x)=\frac{d}{dx}(x + 9)=1 \) (using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), here \( n = 1 \) for \( x \) and derivative of constant 9 is 0).

For \( v(x)=x^2 - 7x + 2 \), using the power rule: \( v'(x)=\frac{d}{dx}(x^2)-\frac{d}{dx}(7x)+\frac{d}{dx}(2)=2x-7 + 0=2x - 7 \).

Step3: Substitute into the quotient rule

Substitute \( u(x), u'(x), v(x), v'(x) \) into the quotient rule formula:

\( f'(x)=\frac{(1)(x^2 - 7x + 2)-(x + 9)(2x - 7)}{(x^2 - 7x + 2)^2} \)

Step4: Expand the numerator

First, expand \( (x + 9)(2x - 7) \):
\( (x + 9)(2x - 7)=x(2x - 7)+9(2x - 7)=2x^2-7x + 18x-63=2x^2 + 11x-63 \)

Now, expand the numerator:
\( (x^2 - 7x + 2)-(2x^2 + 11x-63)=x^2 - 7x + 2-2x^2-11x + 63 \)

Combine like terms:
\( (x^2-2x^2)+(-7x-11x)+(2 + 63)=-x^2-18x + 65 \)

Step5: Write the final derivative

So, \( f'(x)=\frac{-x^2-18x + 65}{(x^2 - 7x + 2)^2} \) (we can also factor out a negative sign from the numerator: \( \frac{-(x^2 + 18x - 65)}{(x^2 - 7x + 2)^2} \), but the expanded form is also correct).

Answer:

\( \frac{-x^2 - 18x + 65}{(x^2 - 7x + 2)^2} \)