keche0b

2021-12-22

How to find f'(0) ?
$f\left(x\right)=\frac{{x}^{2}+3}{2x-1}$

### Answer & Explanation

Ethan Sanders

Explanation:
As $f\left(x\right)=\frac{{x}^{2}+3}{2x-1}$
and using quotient rule,
${f}^{\prime }\left(x\right)=\frac{2x×\left(2x-1\right)-2×\left({x}^{2}+3\right)}{{\left(2x-1\right)}^{2}}$
$=\frac{4{x}^{2}-2x-2{x}^{2}-6}{{\left(2x-1\right)}^{2}}$
$=\frac{2{x}^{2}-2x-6}{{\left(2x-1\right)}^{2}}$
and ${f}^{\prime }\left(0\right)=\frac{2\left({0}^{2}-0-3\right)}{{\left(2×0-1\right)}^{2}}=-\frac{6}{1}=-6$

Debbie Moore

Using the quotient rule:
$\frac{df}{dx}=\frac{\left(2x-1\right)\frac{d}{dx}\left({x}^{2}+3\right)-\left({x}^{2}+3\right)\frac{d}{dx}\left(2x-1\right)}{{\left(2x-1\right)}^{2}}$
$\frac{df}{dx}=\frac{2x\left(2x-1\right)-2\left({x}^{2}+3\right)}{{\left(2x-1\right)}^{2}}$
$\frac{df}{dx}=\frac{4{x}^{2}-2x-2{x}^{2}-6}{{\left(2x-1\right)}^{2}}$
$\frac{df}{dx}=\frac{2{x}^{2}-2x-6}{{\left(2x-1\right)}^{2}}$
and for $x=0$
${\left[\frac{df}{dx}\right]}_{x=0}=-6$

karton

Recall that,

$\begin{array}{}{f}^{\prime }\left(a\right)=\underset{x\to a}{lim}\frac{f\left(x\right)-f\left(a\right)}{x-a}\\ {f}^{\prime }\left(0\right)=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x}\\ {f}^{\prime }\left(0\right)=\underset{x\to 0}{lim}\left[\frac{1}{x}\left(\frac{{x}^{2}+3}{2x-1}-\left(-3\right)\right)\right]\\ =\underset{x\to 0}{lim}\frac{1}{x}\left[\frac{{x}^{2}+3+3\left(2x-1\right)}{2x-1}\right]\\ =\underset{x\to 0}{lim}\frac{1}{x}\frac{{x}^{2}+6x}{2x-1}\\ =\underset{x\to 0}{lim}\frac{x\left(x+6\right)}{2x-1}\\ ⇒{f}^{\prime }\left(0\right)=\frac{0+6}{0-1}=-6\end{array}$

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