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Question Number 78136 by msup trace by abdo last updated on 14/Jan/20

g i v e t h e e q u a t i o n o f t a n g e n t e

a t p (x_{0}, f (x_{0}))

1) f (x) = e^{- x^{2}} l n (1 - 2 x) x_{0} = - 1

2) f (x) = (x^{2} - 3) a r c t a n (x^{2})

x_{0} = 1

3) f (x) = \frac{e^{- x} s i n (π x)}{x^{2} + 3} x_{0} = \frac{1}{2}

4) f (x) = e^{- x} \sqrt{e^{x^{2}} - 1}

a n d x_{0} = 2

Commented by mathmax by abdo last updated on 14/Jan/20

1) e q u . o f t a n g e n t e i s y = f^{^{'}} (- 1) (x + 1)) f (- 1)

f (- 1) = e^{- 1} l n (3) a n d f^{^{'}} (x) = - 2 {x e}^{- x^{2}} l n (1 - 2 x) - \frac{2}{1 - 2 x} e^{- x^{2}} \Rightarrow

f^{^{'}} (- 1) = 2 e^{- 1} l n (3) - \frac{2}{3} e^{- 1} = (2 l n (3) - \frac{2}{3}) e^{- 1} \Rightarrow

y = (2 l n (3) - \frac{2}{3}) (x + 1) + e^{- 1} l n (3)

Commented by mathmax by abdo last updated on 14/Jan/20

1) y = f^{^{'}} (- 1) (x + 1) + f (- 1)

Commented by jagoll last updated on 14/Jan/20

w r o n g t h e f o u r l i n e

Commented by msup trace by abdo last updated on 15/Jan/20

f o r g i v e y = (2 l n 3 - \frac{2}{3}) e^{- 1} (x + 1) + e^{- 1} l n (3)

Commented by msup trace by abdo last updated on 15/Jan/20

2) f (x) = (x^{2} - 3) a t c t a n (x^{2}) \Rightarrow

f (1) = - 2 \times \frac{π}{4} = - \frac{π}{2} a n d

f^{^{'}} (x) = 2 x a r c t a n (x^{2}) + \frac{2 x (x^{2} - 3)}{1 + x^{4}}

\Rightarrow f^{^{'}} (1) = 2 . \frac{π}{4} + \frac{- 4}{2} = \frac{π}{2} - 2 \Rightarrow

t h e r q u a t i o n o f t a n g e n t e i s

y = f^{^{'}} (1) (x - 1) + f (1)

= (\frac{π}{2} - 2) (x - 1) - \frac{π}{2}

= (\frac{π}{2} - 2) x - \frac{π}{2} + 2 - \frac{π}{2} \Rightarrow

y = (\frac{π}{2} - 2) x + 2 - π

Answered by jagoll last updated on 14/Jan/20

(1) f (x_{0}) = e^{- 1} l n (3) = \frac{l n (3)}{e}

s l o p e = m = f^{'} (- 1)

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

f^{'} (- 1) = 2 e^{- 1} l n (3) - \frac{2 e^{- 1}}{3}

= \frac{2 l n (3)}{e} - \frac{2}{3 e} = \frac{6 l n (3) - 2}{3 e}

t a n g e n t l i n e i s

y = \frac{6 l n (3) - 2}{3 e} (x + 1) + \frac{l n (3)}{e}

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