Question

Sa se determine f○g si g○f

f:R-R
f(x) = 2x-1 , x >sau eg 3
-2x+3 , x<3

g:R-R

g(x)=x patrat - 1​

Answer 1

Explicație pas cu pas:

$f(x) = \begin{cases} \ \ \ 2x - 1, \ x \geqslant 3 \\ - 2x + 3 , \ x < 3\end{cases}$

$g(x) = {x}^{2} - 1, \ x \in \mathbb{R}$

$(f ○ g)(x) = f(g(x)) = \begin{cases} \ \ \ 2 \cdot g(x) - 1, \ g(x) \geqslant 3\\ - 2 \cdot g(x) + 3 , \ g(x) < 3\end{cases}$

$= \begin{cases} \ \ \ 2( {x}^{2} - 1) - 1, \ {x}^{2} - 1 \geqslant 3, \ x \geqslant 3 \\ - 2( {x}^{2} - 1) + 3 , \ {x}^{2} - 1 < 3, \ x < 3\end{cases}$

$= \begin{cases} \ \ \ 2 {x}^{2} - 3, \ {x}^{2} \geqslant 4\\ - 2 {x}^{2} + 5, \ {x}^{2} < 4\end{cases}$

$= \begin{cases} \ \ \ 2 {x}^{2} - 3, \ x \in \Big(-\infty ; -2 \Big] ∪ \Big[2 ; +\infty \Big)\\ - 2 {x}^{2} + 5 , \ x \in \Big(-2; 2 \Big)\end{cases}$

$(g ○ f)(x) = g(f(x)) = {(f(x))}^{2} - 1 =$

$= \begin{cases} {(f(x))}^{2} - 1, \ x \geqslant 3 \\ {(f(x))}^{2} - 1, \ x < 3\end{cases}$

$= \begin{cases} \ \ \ {(2x - 1)}^{2} - 1, \ x \geqslant 3 \\ {( - 2x + 3)}^{2} - 1 , \ x < 3\end{cases}$

$= \begin{cases} \ \ \ \ \ \ \ \ \ 4 {x}^{2} - 4x, \ x \geqslant 3 \\ 4 {x}^{2} - 12x + 8 , \ x < 3\end{cases}$

$= \begin{cases} \ \ \ \ \ \ \ \ \ 4x(x - 1), \ x \in \Big[3 ; +\infty \Big) \\ 4 (x - 1)(x - 2), \ x \in \Big( - \infty;3 \Big) \Big)\end{cases}$