What is the equation of the tangent line of f(x)=(x−1)2x+2 at x=3?

Explanation:mtangent=f'(x)&nbsp;at&nbsp;x=adifferentiate using the quotient rulegiven&nbsp;f(x)=g(x)h(x)&nbsp;thenf′(x)=h(x)g′(x)−g(x)h′(x)(h(x))2←&nbsp;quotient ruleg(x)=(x−1)2⇒g′(x)=2(x−1)h(x)=x+2⇒g′(x)=2(x−1)⇒f′(x)=(x+2)2(x−1)−(x−1)2(x+2)2⇒f′(3)=20−425=1625and&nbsp;f(3)=45using&nbsp;m=1625&nbsp;and&nbsp;(x1,y1)=(3,45)⇒y−45=1625(x−3)⇒y=1625x−2825←&nbsp;equation of tangent

Key to "What is the equation of the tangent line..."

Colten Welch

Answered question

2022-04-07

What is the equation of the tangent line of

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

at x=3?

Answer & Explanation

Kapovkah7hpl

Beginner2022-04-08Added 14 answers

Explanation:
$m_{tangent} = f^{'} (x)$ at $x = a$
differentiate using the quotient rule
given $f (x) = \frac{g (x)}{h (x)}$ then
$f^{'} (x) = \frac{h (x) g^{'} (x) - g (x) h^{'} (x)}{{(h (x))}^{2}} \leftarrow$ quotient rule
$g (x) = {(x - 1)}^{2} \Rightarrow g^{'} (x) = 2 (x - 1)$
$h (x) = x + 2 \Rightarrow g^{'} (x) = 2 (x - 1)$
$\Rightarrow f^{'} (x) = \frac{(x + 2) 2 (x - 1) - {(x - 1)}^{2}}{{(x + 2)}^{2}}$
$\Rightarrow f^{'} (3) = \frac{20 - 4}{25} = \frac{16}{25}$
and $f (3) = \frac{4}{5}$
using $m = \frac{16}{25}$ and $(x_{1}, y_{1}) = (3, \frac{4}{5})$
$\Rightarrow y - \frac{4}{5} = \frac{16}{25} (x - 3)$
$\Rightarrow y = \frac{16}{25} x - \frac{28}{25} \leftarrow$ equation of tangent

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