what-is-the-value-of-x-if-f-x-1-x-2-1-g-x-2x-7-and-f-g-1-x-3-

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)

\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