Question

100 de puncte! exercitiul 1 si 2. multumesc!​

Answer 1

$\it 2)\ (f\circ f)(x)=f(f(x))=1-f(x)=1-(1-x)=1-1+x=x\\ \\ \\ 1)\ (f\circ f)(-1)=f(f(-1))=f(1+1-1)=f(1)=1\\ \\ (g\circ g)(1)=g(g(1))=g(2-3)=g(-1)=-2-3=-5\\ \\ (f\circ g)(0)=f(g(0))=f(-3)=9-3-1=5$

Answer 2

Răspuns:

nu stiu daca ti le-am facut pe toate,dar ti-am explicat modelul

Explicație pas cu pas:

2

(f°f)(x)=f(1-x) =1-(1-x) =1-1+x=x

exe 1

f(x) =x²+x-1

g(x) =2x-3

h(x)=3x+1

dupa cum am arata la 2, (f°f)(x) =x, deci(f°f)(-1)=-1

(g°g) (1) =g(2*1-3) =g(-1)=-2-3=-5

(g°h)(-2)=g(h(-2))=g(-6+1)=g(-5)=-10-3=-13

(f°g)(0)=f(g(0))=f(-3)=9-5-1=5

(h°h)(f(1))=(h°h)(1+1-1) =(h°h)(1) =h(h(1))=h(3*1+1) =h(4)=3*4+1=13