find f(x) and find the value(s) of x where the tangent line is horizontal. f(x) = x / (2x - 7)^4 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 $v=(2x - 7)^{4}$, so we first find $v'$ using the chain - rule. Let $t = 2x-7$, then $v=t^{4}$, and $\frac{dv}{dt}=4t^{3}$ and $\frac{dt}{dx}=2$. So $v' = 4(2x - 7)^{3}\times2=8(2x - 7)^{3}$.

Step2: Calculate $f'(x)$

$f'(x)=\frac{1\times(2x - 7)^{4}-x\times8(2x - 7)^{3}}{(2x - 7)^{8}}$. Factor out $(2x - 7)^{3}$ from the numerator: $f'(x)=\frac{(2x - 7)^{3}[(2x - 7)-8x]}{(2x - 7)^{8}}=\frac{2x - 7-8x}{(2x - 7)^{5}}=\frac{-6x - 7}{(2x - 7)^{5}}$.

Step3: Find when tangent is horizontal

The tangent line is horizontal when $f'(x)=0$. A fraction $\frac{a}{b}=0$ when $a = 0$ and $b\neq0$. Set the numerator equal to 0: $-6x-7 = 0$. Solve for $x$: $x=-\frac{7}{6}$.

Answer:

$f'(x)=\frac{-6x - 7}{(2x - 7)^{5}}$; The value of $x$ where the tangent line is horizontal is $x =-\frac{7}{6}$