# Thread: In each case compute f o g (x) where

1. ## In each case compute f o g (x) where

f(x) = 2x + 3 and g(x) = x^2 − 1.

fog(x)= 2(x^2-1)+3
gof(x) = 2x^2+1

gof (x) dosent equal to fog(x)

f(x) = x^2 − 1 and g(x) = 2x + 3.

fog ( x) = x(2x+3)^2-1
gof(x)= 4x^2+12x+8

gof (x) dosent equal to fog(x)

are thses right with proper working ?

if not can some1 show me the answear and the proper working
cheers

2. ## Re: In each case compute f o g (x) where

Originally Posted by arsenal12345
f(x) = 2x + 3 and g(x) = x^2 − 1.

fog(x)= 2(x^2-1)+3
gof(x) = 2x^2+1

gof (x) dosent equal to fog(x)

f(x) = x^2 − 1 and g(x) = 2x + 3.

fog ( x) = x(2x+3)^2-1
gof(x)= 4x^2+12x+8

gof (x) dosent equal to fog(x)

are thses right with proper working ?

if not can some1 show me the answear and the proper working
cheers
In question 1, \displaystyle \begin{align*} f \circ g (x) \end{align*} is correct, \displaystyle \begin{align*} g \circ f(x) \end{align*} is not.

3. ## Re: In each case compute f o g (x) where

Originally Posted by Prove It
In question 1, \displaystyle \begin{align*} f \circ g (x) \end{align*} is correct, \displaystyle \begin{align*} g \circ f(x) \end{align*} is not.
is it 4x^3 + 9x+12 - 1 ??

fog (x) dosent eqauel to gof (x)

4. ## Re: In each case compute f o g (x) where

Originally Posted by arsenal12345
is it 4x^3 + 9x+12 - 1 ??

fog (x) dosent eqauel to gof (x)
No.

\displaystyle \begin{align*} g \circ f(x) &= g\left(f(x)\right) \\ &= \left(2x + 3\right)^2 - 1 \\ &= 4x^2 + 12x + 9 - 1 \\ &= 4x^2 + 12x + 8 \end{align*}