Given-a-function-y-f-x-where-f-1-x-5-x-5-8-x-5-Find-slope-of-the-curve-y-f-x-at-x-1-

Question Number 124007 by liberty last updated on 30/Nov/20

G i v e n a f u n c t i o n y = f (x) w h e r e f^{- 1} (\frac{x + 5}{x - 5}) = \frac{8}{x + 5}

F i n d s l o p e o f t h e c u r v e y = f (x) a t x = 1 .

Answered by john_santu last updated on 30/Nov/20

f^{- 1} (\frac{x + 5}{x - 5}) = \frac{8}{x + 5} \Leftrightarrow f (\frac{8}{x + 5}) = \frac{x + 5}{x - 5}

d i f f e r e n t i a t i n g b o t h s i d e g i v e

- \frac{8}{{(x + 5)}^{2}} f^{'} (\frac{8}{x + 5}) = - \frac{10}{{(x - 5)}^{2}}

p u t x = 3 \Rightarrow - \frac{8}{64} f^{'} (1) = - \frac{10}{4}

\Rightarrow f^{'} (1) = \frac{10}{4} \times \frac{8}{1} = 20 .

Answered by mathmax by abdo last updated on 30/Nov/20

f^{- 1} (\frac{x + 5}{x - 5}) = \frac{8}{x + 5} \Rightarrow f (\frac{8}{x + 5}) = \frac{x + 5}{x - 5} let \frac{8}{x + 5} = t \Rightarrow tx + 5 t = 8 \Rightarrow

tx = 8 - 5 t \Rightarrow x = \frac{8 - 5 t}{t} \Rightarrow f (t) = \frac{\frac{8 - 5 t}{t} + 5}{\frac{8 - 5 t}{t} - 5} = \frac{8 t}{8 - 10 t} = \frac{4 t}{4 - 5 t}

we have f (t) = \frac{4 t}{4 - 5 t} \Rightarrow f^{^{'}} (t) = \frac{4 (4 - 5 t) - 4 t (- 5)}{{(4 - 5 t)}^{2}} = \frac{16 - 20 t + 20 t}{{(5 t - 4)}^{2}}

\Rightarrow f^{^{'}} (t) = \frac{16}{{(5 t - 4)}^{2}} \Rightarrow f^{^{'}} (1) = 16

Commented by liberty last updated on 30/Nov/20

n o t c o r r e c t .

f (t) = \frac{\frac{8 - 5 t}{t} + 5}{\frac{8 - 5 t}{t} - 5} = \frac{8 - 5 t + 5 t}{8 - 5 t - 5 t} = \frac{8}{8 - 10 t} = \frac{4}{4 - 5 t}

s i r =

Commented by Bird last updated on 30/Nov/20

s o r r y f (t) = \frac{8}{8 - 10 t} = \frac{4}{4 - 5 t} \Rightarrow

f^{^{'}} (t) = 4 \times \frac{5}{{(4 - 5 t)}^{2}} = \frac{20}{{(4 - 5 t)}^{2}} \Rightarrow

s l o p e = f^{^{'}} (1) = 20

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