# What is the instantaneous rate of change of f(x)=1/(x^2-x+3 ) at x=0 ?

Jan 20, 2018

1/9

#### Explanation:

find $f ' \left(x\right)$ or the instantaneous rate of change of $f \left(x\right)$ at x.

$= - \frac{1}{{\left({x}^{2} - x + 3\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - x + 3\right)$

(power rule: $\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ and chain rule: $\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) g ' \left(x\right)$)

$= - \frac{1}{{\left({x}^{2} - x + 3\right)}^{2}} \cdot \left(2 x - 1\right)$

plug in 0 for x:

$f ' \left(0\right) = - \frac{1}{{\left({0}^{2} - 0 + 3\right)}^{2}} \cdot \left(2 \left(0\right) - 1\right)$

$f ' \left(0\right) = - \frac{1}{{3}^{2}} \cdot \left(- 1\right)$

$f ' \left(0\right) = \frac{1}{9}$

Jan 20, 2018

$\frac{1}{9}$

#### Explanation:

$\text{the instantaneous rate of change of f(x) at x=0}$
$\text{is f'(0)}$

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

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

$f ' \left(x\right) = g ' \left(h \left(x\right)\right) \times h ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{chain rule}}$

$f \left(x\right) = \frac{1}{{x}^{2} - x + 3} = {\left({x}^{2} - x + 3\right)}^{-} 1$

$\Rightarrow f ' \left(x\right) = - {\left({x}^{2} - x + 3\right)}^{-} 2 \times \frac{d}{\mathrm{dx}} \left({x}^{2} - x + 3\right)$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = - \frac{2 x - 1}{{x}^{2} - x + 3} ^ 2$

$\Rightarrow f ' \left(0\right) = - \frac{- 1}{3} ^ 2 = \frac{1}{9}$