what is the value of x if f(x+1) = x^2 −1 g(x)= 2x+7 and f(g^(−1) (x))= 3


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Question Number 94366 by i jagooll last updated on 18/May/20

$what is the value of x if f (x + 1) = x^{2} - 1$ $g (x) = 2 x + 7 and f (g^{- 1} (x)) = 3$
Commented by mr W last updated on 18/May/20

$g^{- 1} (x) = \frac{x - 7}{2}$ $f (x) = {(x - 1)}^{2} - 1$ $f (g^{- 1} (x)) = {(\frac{x - 7}{2} - 1)}^{2} - 1 = 3$ $\frac{x - 7}{2} - 1 = \pm 2$ $\Rightarrow x = 13 o r 5$
Commented by i jagooll last updated on 18/May/20

$thank you both$
Commented by Abdulrahman last updated on 18/May/20

$please$ $solve$ $this$ $3^{x} + x^{3} = 17$ $with$ $steps$
Commented by mr W last updated on 18/May/20

$1 . w e s e e (i f y o u {d o n}^{'} t s e e, t h e n i {c a n}^{'} t$ $h e l p y o u), t h a t x = 2 i s a r o o t, s i n c e$ $3^{2} + 2^{3} = 9 + 8 = 17 .$ $2 . b o t h 3^{x} a n d x^{3} a r e s t r i c t l y i n c r e a s i n g$ $f u n c t i o n s, s o 3^{x} + x^{3} i s a l s o s t r i c t l y$ $i n c r e a s i n g . w h e n a s t r i c t l y i n c r e a s i n g$ $f u n c t i o n h a s a z e r o, t h e n t h i s z e r o$ $i s t h e o n l y o n e z e r o . t h a t m e a n s$ $x = 2 i s t h e o n l y o n e r o o t o f e q n .$ $x^{2} + 3^{x} = 17 .$
Commented by Abdulrahman last updated on 18/May/20

$bundle of thanks$
Commented by Abdulrahman last updated on 18/May/20

$\int \frac{{(x - 2)}^{3}}{2 - x^{2}} dx = ?$
Commented by abdomathmax last updated on 18/May/20

$I = \int \frac{{(x - 2)}^{3}}{2 - x^{2}} dx \Rightarrow I = \int \frac{x^{3} - {3 x}^{2} \times 2 + 3 x \times 2^{2} - 2^{3}}{2 - x^{2}} dx$ $= \int \frac{x^{3} - {6 x}^{2} + 12 x + 8}{2 - x^{2}} dx = - \int \frac{x^{3} - {6 x}^{2} + 12 x + 8}{x^{2} - 2} dx$ $= - \int \frac{x (x^{2} - 2) - {6 x}^{2} + 14 x + 8}{x^{2} - 2} dx$ $= - \int xdx + \int \frac{{6 x}^{2} - 14 x + 8}{x^{2} - 2} dx$ $= - \frac{x^{2}}{2} + \int \frac{6 (x^{2} - 2) - 14 x + 20}{x^{2} - 2} dx$ $= - \frac{x^{2}}{2} + 6 x - 2 \int \frac{7 x - 10}{x^{2} - 2} dx$ $F (x) = \frac{7 x - 10}{x^{2} - 2} = \frac{7 x - 10}{(x - \sqrt{2}) (x + \sqrt{2})} = \frac{a}{x - \sqrt{2}} + \frac{b}{x + \sqrt{2}}$ $a = \frac{7 \sqrt{2} - 10}{2 \sqrt{2}} and b = \frac{- 7 \sqrt{2} - 10}{- 2 \sqrt{2}} = \frac{7 \sqrt{2} + 10}{2 \sqrt{2}}$ $\int \frac{7 x - 10}{x^{2} - 2} dx = \frac{7 \sqrt{2} - 10}{2 \sqrt{2}} \ln ∣ x - \sqrt{2} ∣ + \frac{7 \sqrt{2} + 10}{2 \sqrt{2}} \ln ∣ x + \sqrt{2} ∣ + c$ $I = - \frac{x^{2}}{2} + 6 x - (7 \sqrt{2} - 10) \ln ∣ x - \sqrt{2} ∣ - (7 \sqrt{2} + 10) \ln ∣ x + \sqrt{2} ∣ + C$
Commented by mr W last updated on 18/May/20

$a b d u l r a m a n s i r :$ $p l e a s e o p e n a n e w t h r e a d i f y o u w a n t$ $t o a s k a n e w q u e s t i o n!$
Commented by Abdulrahman last updated on 18/May/20

$ok thanks alot$
	Answered by john santu last updated on 18/May/20
	$g^(−1) (x) = f^(−1) (3) ⇒f^(−1) (x^2 −1) = x+1 ⇒ { ((x^2 −1= 3)),((⇒ { ((x = 2)),((x = −2)) :})) :} case(1) f^(−1) (3)= 3 & g^(−1) (x)=((x−7)/2) ⇒ ((x−7)/2) = 3 ⇒ x = 13 case(2) f^(−1) (3)=−1 & g^(−1) (x)=((x−7)/2) ⇒ ((x−7)/2) = −1 ⇒ x = 5 solution x = 5 or 13$
	$g^{- 1} (x) = f^{- 1} (3)$ $\Rightarrow f^{- 1} (x^{2} - 1) = x + 1 \Rightarrow {\begin{cases} x^{2} - 1 = 3 \\ \Rightarrow {\begin{cases} x = 2 \\ x = - 2 \end{cases} \end{cases}$ $case (1) f^{- 1} (3) = 3 & g^{- 1} (x) = \frac{x - 7}{2}$ $\Rightarrow \frac{x - 7}{2} = 3 \Rightarrow x = 13$ $case (2) f^{- 1} (3) = - 1 & g^{- 1} (x) = \frac{x - 7}{2}$ $\Rightarrow \frac{x - 7}{2} = - 1 \Rightarrow x = 5$ $solution x = 5 or 13$
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