given: $f(x)=-\frac{5x + 3}{2x - 2}$. find $f(x)$. a $f(x)=\frac{3x + 5}{4x^{2}-8x + 4}$ b $f(x)=-\frac{5x…

given: $f(x)=-\frac{5x + 3}{2x - 2}$. find $f(x)$. a $f(x)=\frac{3x + 5}{4x^{2}-8x + 4}$ b $f(x)=-\frac{5x - 1}{x^{2}-2x + 1}$ c $f(x)=\frac{4}{x^{2}-2x + 1}$ d $f(x)=-\frac{7x + 1}{4x^{2}-8x + 4}$ e $f(x)=-\frac{21x + 19}{4x^{2}-8x + 4}$

given: $f(x)=-\frac{5x + 3}{2x - 2}$. find $f(x)$. a $f(x)=\frac{3x + 5}{4x^{2}-8x + 4}$ b $f(x)=-\frac{5x - 1}{x^{2}-2x + 1}$ c $f(x)=\frac{4}{x^{2}-2x + 1}$ d $f(x)=-\frac{7x + 1}{4x^{2}-8x + 4}$ e $f(x)=-\frac{21x + 19}{4x^{2}-8x + 4}$

Answer

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. For $f(x)=-\frac{5x + 3}{2x - 2}$, let $u=-(5x + 3)=-5x-3$ and $v = 2x - 2$. Then $u'=-5$ and $v'=2$.

Step2: Substitute into quotient - rule formula

$f'(x)=\frac{(-5)(2x - 2)-(-5x - 3)\times2}{(2x - 2)^{2}}$. First, expand the numerator: $(-5)(2x - 2)-(-5x - 3)\times2=-10x + 10+10x + 6$. Combine like - terms in the numerator: $(-10x+10x)+(10 + 6)=16$. The denominator is $(2x - 2)^{2}=4x^{2}-8x + 4$. So $f'(x)=\frac{16}{4x^{2}-8x + 4}=\frac{4}{x^{2}-2x + 1}$.

Answer:

C. $f'(x)=\frac{4}{x^{2}-2x + 1}$