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Question Number 80180 by mathocean1 last updated on 31/Jan/20

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

w e d e f i n e d t h i s f u n c t i o n o n

R - {- 1} \to R

1) Study the variations of h then

draw up its table of variation .

please sirs i need your kind help

Answered by Rio Michael last updated on 31/Jan/20

solution

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

h (x) = 0 \Rightarrow x - x^{2} = 0 \Rightarrow x = 0 o r x = 1

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

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

h^{^{'}} (x) > 0 \Rightarrow {(x - 1)}^{2} > 0 \Rightarrow x > 1

h i s i n c r e a s i n g f o r x > 1 a n d d e c r e a s i n g o t h e r w i s e

Commented by mathocean1 last updated on 31/Jan/20

Please sir h^{'} is not like this :

h^{'} = \frac{(1 - 2 x) (x + 1) - (x - x^{2})}{{(x + 1)}^{2}} ?

Commented by mathocean1 last updated on 31/Jan/20

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

Commented by Rio Michael last updated on 31/Jan/20

s o r r y i c o p i e d t h e q u e s t i o n w r o n g l y

Commented by mathocean1 last updated on 31/Jan/20

please sir how can i draw up its

table of variation when

h^{'} = \frac{- x^{2} - 2 x + 1}{{(x + 1)}^{2}} = - 1 ?

Commented by JDamian last updated on 01/Feb/20

i f h (x) w e r e \frac{x - x^{2}}{x - 1}, t h e n h (x) = \frac{x (1 - x)}{x - 1} = - x

a v e r y e a s y f u n c t i o n

Commented by mathocean1 last updated on 01/Feb/20

so \dots we can study h (x) = \frac{x - x^{2}}{x + 1} ? ? ?