pevetaty

2022-03-15

What is the instantaneous rate of change of $f\left(x\right)=\frac{1}{{x}^{2}-x+3}$ at x=0?

Raster02f

Find f'(x) or the instantaneous rate of change of f(x) at x.
$=-\frac{1}{{\left({x}^{2}-x+3\right)}^{2}}\cdot \frac{d}{dx}\left({x}^{2}-x+3\right)$
(power rule: $\frac{d}{dx}\left({x}^{n}\right)=n{x}^{n-1}$ and chain rule: $\frac{d}{dx}\left(f\left(g\left(x\right)\right)\right)={f}^{\prime }\left(g\left(x\right)\right){g}^{\prime }\left(x\right)\right)$
$=-\frac{1}{{\left({x}^{2}-x+3\right)}^{2}}\cdot \left(2x-1\right)$
plug in 0 for x:
${f}^{\prime }\left(0\right)=-\frac{1}{{\left({0}^{2}-0+3\right)}^{2}}\cdot \left(2\left(0\right)-1\right)$
${f}^{\prime }\left(0\right)=-\frac{1}{{3}^{2}}\cdot \left(-1\right)$
${f}^{\prime }\left(0\right)=\frac{1}{9}$

Ciara Hoffman

Explanation:
the instantaneous rate of change of f(x) at x=0 is f'(0)
differentiate using the chain rule
given f(x)=g(h(x)) then
${f}^{\prime }\left(x\right)={g}^{\prime }\left(h\left(x\right)\right)×{\left({x}^{2}-x+3\right)}^{-1}$
$⇒{f}^{\prime }\left(x\right)=-{\left({x}^{2}-x+3\right)}^{-2}×\frac{d}{dx}\left({x}^{2}-x+3\right)$
$=-\frac{2x-1}{{\left({x}^{2}-x+3\right)}^{2}}$
$⇒{f}^{\prime }\left(0\right)=-\frac{-1}{{3}^{2}}=\frac{1}{9}$

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