Question

find f(x) and find the value(s) of x where the tangent line is horizontal.
f(x) = \frac{x}{(6x - 5)^3}
f(x) =

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x$, so $u'=1$, and $v=(6x - 5)^{3}$, so $v'=3(6x - 5)^{2}\times6 = 18(6x - 5)^{2}$.

Step2: Calculate $f'(x)$

$f'(x)=\frac{1\times(6x - 5)^{3}-x\times18(6x - 5)^{2}}{(6x - 5)^{6}}$.
Factor out $(6x - 5)^{2}$ from the numerator:
$f'(x)=\frac{(6x - 5)^{2}[(6x - 5)-18x]}{(6x - 5)^{6}}$.
Simplify the numerator: $(6x - 5)-18x=- 12x - 5$.
So, $f'(x)=\frac{-12x - 5}{(6x - 5)^{4}}$.

Step3: Find when the tangent is horizontal

The tangent line is horizontal when $f'(x)=0$. For a fraction $\frac{a}{b}$ to be $0$, the numerator $a$ must be $0$ (while $b
eq0$).
Set $-12x - 5 = 0$.
Solve for $x$: $-12x=5$, so $x=-\frac{5}{12}$. And when $x = -\frac{5}{12}$, the denominator $(6x - 5)^{4}
eq0$.

Answer:

$f'(x)=\frac{-12x - 5}{(6x - 5)^{4}}$; $x = -\frac{5}{12}$