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# What is the instantaneous rate of change of f(x)=x/(-x-8) at x=4 ?

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Feb 9, 2018

$\setminus$

$\text{Answer is:} \setminus q \quad \setminus \quad - \frac{1}{18.}$

#### Explanation:

$\setminus$

$\text{The required quantity is the derivative of" \ \ f(x) \ \ "evaluated at}$
$\setminus q \quad \setminus q \quad x = 4.$

$\text{So, let's calculate the derivative of} \setminus \setminus f \left(x\right) :$

1. $\text{Original Function:} \setminus q \quad f \left(x\right) = \frac{x}{- x - 8}$
2. $\text{Rewrite a little:} \setminus q \quad f \left(x\right) = - \frac{x}{x + 8}$
3. $\text{Quotient Rule:} \setminus q \quad f ' \left(x\right) = - \frac{\left(x + 8\right) \left[x\right] ' - x \left[x + 8\right] '}{{\left(x + 8\right)}^{2}}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus = - \frac{\left(x + 8\right) \left[1\right] - x \left[1 + 0\right]}{{\left(x + 8\right)}^{2}}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus = - \frac{\left(x + 8\right) - x}{{\left(x + 8\right)}^{2}}$
$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus = - \frac{8}{{\left(x + 8\right)}^{2}} .$
4. $\text{Summary:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus f ' \left(x\right) = - \frac{8}{{\left(x + 8\right)}^{2}} .$
5. $\text{Evaluate" \ \ f'(x) \ \ "at} \setminus \setminus x = 4 ,$
$\setminus q \quad \setminus \quad \setminus \quad \text{to get the instanteous rate of change at} \setminus \setminus x = 4 :$
$f ' \left(4\right) = - \frac{8}{{\left(4 + 8\right)}^{2}} \setminus = - \setminus \frac{8}{12} ^ 2 \setminus = - \setminus \frac{8}{12 \setminus \cdot 12}$
$\setminus q \quad \setminus \quad \setminus \quad \setminus = - \setminus \frac{\textcolor{red}{\cancel{8}}}{\textcolor{red}{\cancel{2}} \setminus \cdot 6 \setminus \cdot \textcolor{red}{\cancel{4}} \setminus \cdot 3} = \setminus \setminus - \frac{1}{18.}$
6. $\text{Summarize:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \quad \setminus f ' \left(4\right) = - \frac{1}{18.}$

$\therefore \text{instantaneous rate of change of"\ \ f(x) \ \ "at} \setminus \left(x = 4\right) \setminus = - \frac{1}{18.}$

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Jim G. Share
Feb 9, 2018

$- \frac{1}{18}$

#### Explanation:

$\text{the instantaneous rate of change at x = 4}$

$\text{is the value of the derivative at x = 4}$

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \textcolor{b l u e}{\text{quotient rule}}$

$g \left(x\right) = x \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = - x - 8 \Rightarrow h ' \left(x\right) = - 1$

$\Rightarrow f ' \left(x\right) = \frac{- x - 8 - x \left(- 1\right)}{- x - 8} ^ 2 = - \frac{8}{- x - 8} ^ 2$

$\Rightarrow f ' \left(4\right) = - \frac{8}{144} = - \frac{1}{18}$

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