How to find f'(0) ?f(x)=x2+32x−1

Explanation:As f(x)=x2+32x−1and using quotient rule,f′(x)=2x×(2x−1)−2×(x2+3)(2x−1)2=4x2−2x−2x2−6(2x−1)2=2x2−2x−6(2x−1)2and f′(0)=2(02−0−3)(2×0−1)2=−61=−6

Using the quotient rule: dfdx=(2x−1)ddx(x2+3)−(x2+3)ddx(2x−1)(2x−1)2 dfdx=2x(2x−1)−2(x2+3)(2x−1)2 dfdx=4x2−2x−2x2−6(2x−1)2 dfdx=2x2−2x−6(2x−1)2 and for x=0 [dfdx]x=0=−6

Recall that,f′(a)=limx→af(x)−f(a)x−af′(0)=limx→0f(x)−f(0)x−0=limx→0f(x)−f(0)xf′(0)=limx→0[1x(x2+32x−1−(−3))]=limx→01x[x2+3+3(2x−1)2x−1]=limx→01xx2+6x2x−1=limx→0x(x+6)2x−1⇒f′(0)=0+60−1=−6

Explanation:As f(x)=x2+32x−1and using quotient rule,f′(x)=2x×(2x−1)−2×(x2+3)(2x−1)2=4x2−2x−2x2−6(2x−1)2=2x2−2x−6(2x−1)2and f′(0)=2(02−0−3)(2×0−1)2=−61=−6

Using the quotient rule: dfdx=(2x−1)ddx(x2+3)−(x2+3)ddx(2x−1)(2x−1)2 dfdx=2x(2x−1)−2(x2+3)(2x−1)2 dfdx=4x2−2x−2x2−6(2x−1)2 dfdx=2x2−2x−6(2x−1)2 and for x=0 [dfdx]x=0=−6

Recall that,f′(a)=limx→af(x)−f(a)x−af′(0)=limx→0f(x)−f(0)x−0=limx→0f(x)−f(0)xf′(0)=limx→0[1x(x2+32x−1−(−3))]=limx→01x[x2+3+3(2x−1)2x−1]=limx→01xx2+6x2x−1=limx→0x(x+6)2x−1⇒f′(0)=0+60−1=−6

Explained: "How to find f'(0) ?f(x)=x2+32x−1"

keche0b

Answered question

2021-12-22

How to find f'(0) ?

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

Answer & Explanation

Ethan Sanders

Beginner2021-12-23Added 35 answers

Explanation:
As

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

and using quotient rule,

f^{'} (x) = \frac{2 x \times (2 x - 1) - 2 \times (x^{2} + 3)}{{(2 x - 1)}^{2}}

= \frac{4 x^{2} - 2 x - 2 x^{2} - 6}{{(2 x - 1)}^{2}}

= \frac{2 x^{2} - 2 x - 6}{{(2 x - 1)}^{2}}

and

f^{'} (0) = \frac{2 (0^{2} - 0 - 3)}{{(2 \times 0 - 1)}^{2}} = - \frac{6}{1} = - 6

Debbie Moore

Beginner2021-12-24Added 43 answers

Using the quotient rule:

\frac{d f}{d x} = \frac{(2 x - 1) \frac{d}{d x} (x^{2} + 3) - (x^{2} + 3) \frac{d}{d x} (2 x - 1)}{{(2 x - 1)}^{2}}

\frac{d f}{d x} = \frac{2 x (2 x - 1) - 2 (x^{2} + 3)}{{(2 x - 1)}^{2}}

\frac{d f}{d x} = \frac{4 x^{2} - 2 x - 2 x^{2} - 6}{{(2 x - 1)}^{2}}

\frac{d f}{d x} = \frac{2 x^{2} - 2 x - 6}{{(2 x - 1)}^{2}}

and for

x = 0

{[\frac{d f}{d x}]}_{x = 0} = - 6

karton

Expert2021-12-30Added 613 answers

Recall that,

$\begin{matrix} f^{'} (a) = lim_{x \to a} \frac{f (x) - f (a)}{x - a} \\ f^{'} (0) = lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} = lim_{x \to 0} \frac{f (x) - f (0)}{x} \\ f^{'} (0) = lim_{x \to 0} [\frac{1}{x} (\frac{x^{2} + 3}{2 x - 1} - (- 3))] \\ = lim_{x \to 0} \frac{1}{x} [\frac{x^{2} + 3 + 3 (2 x - 1)}{2 x - 1}] \\ = lim_{x \to 0} \frac{1}{x} \frac{x^{2} + 6 x}{2 x - 1} \\ = lim_{x \to 0} \frac{x (x + 6)}{2 x - 1} \\ \Rightarrow f^{'} (0) = \frac{0 + 6}{0 - 1} = - 6 \end{matrix}$

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