find f(x) and find the value(s) of x where the tangent line is horizontal.\nf(x)=\frac{x}{(6x - 5)^4}\nf(x)=□

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x$, so $u^\prime=1$, and $v=(6x - 5)^{4}$, so $v^\prime = 4(6x - 5)^{3}\times6=24(6x - 5)^{3}$.

Step2: Calculate $f^\prime(x)$

[ \begin{align*} f^\prime(x)&=\frac{1\times(6x - 5)^{4}-x\times24(6x - 5)^{3}}{(6x - 5)^{8}}\ &=\frac{(6x - 5)^{3}[(6x - 5)-24x]}{(6x - 5)^{8}}\ &=\frac{6x - 5-24x}{(6x - 5)^{5}}\ &=\frac{- 18x - 5}{(6x - 5)^{5}} \end{align*} ]

Step3: Find when tangent is horizontal

The tangent line is horizontal when $f^\prime(x)=0$. So we set $\frac{-18x - 5}{(6x - 5)^{5}}=0$. A fraction is zero when the numerator is zero and the denominator is non - zero. Set $-18x - 5 = 0$. Then $18x=-5$, so $x =-\frac{5}{18}$. The denominator $(6x - 5)^{5}\neq0$ when $x =-\frac{5}{18}$.

Answer:

$f^\prime(x)=\frac{-18x - 5}{(6x - 5)^{5}}$, and the value of $x$ where the tangent line is horizontal is $x =-\frac{5}{18}$