Question

Помогите 1.95 срочно

Answer 1

$1)\ \ f(x)=x^2\ ,\ \ g(x)=2x-5\\\\f(g(x))=(g(x))^2=(2x-5)^2\\\\g(f(x))=2\cdot f(x)-5=2\, x^2-5\\\\\\2)\ \ f(x)=x^2+1\ \ ,\ \ g(x)=\sqrt{3-x}\\\\f(g(x))=(g(x))^2+1=(3-x)+1=4-x\\\\g(f(x))=\sqrt{3-f(x)}=\sqrt{3-(x^2+1)}=\sqrt{2-x^2}\\\\\\3)\ \ f(x)=x^2-1\ \ ,\ \ g(x)=x^2+1\\\\f(g(x))=(g(x))^2-1=(x^2+1)^2-1=x^4+2x^2\\\\g(f(x))=(f(x))^2+1=(x^2-1)^2-1=x^4-2x^2$

$4)\ \ f(x)=\dfrac{2}{x}\ \ ,\ \ g(x)=x^2-3x+2\\\\f(g(x))=\dfrac{2}{g(x)}=\dfrac{2}{x^2-3x+2}\\\\\\g(f(x))=f^2(x)-3\cdot f(x)+2=\dfrac{4}{x^2}-\dfrac{6}{x}+2$