Question

Запишите сложные функции f(g(x)) и g(f(x)) 1)f(x)=x^2 и g(x)=2x-52)f(x)=x^2+1 и g(x)=√3-x3)f(x)=x^2-1 и g(x)=x^2+14)f(x)=2/x и g(x)=x^2-3x+2​

Answer 1

$1)\ \ f(x)=x^2\ ,\ \ g(x)=2x-5\\\\f(g(x))=(2x-5)^2\ \ ,\ \ \ g(f(x))=2x^2-5\\\\\\2)\ \ f(x)=x^2+1\ \ ,\ \ g(x)=\sqrt{3-x}\\\\\\f(g(x))=(3-x)+1=4-x\ \ ,\ \ \ g(f(x))=\sqrt{3-(x^2+1)}=\sqrt{2-x^2}\\\\\\3)\ \ f(x)=x^2-1\ \ ,\ \ g(x)=x^2+1\\\\f(g(x))=(x^2+1)^2-1=x^4+2x^2\ \ ,\ \ g(f(x))=(x^2-1)^2+1=x^4-2x^2+2\\\\\\4)\ \ f(x)=\dfrac{2}{x}\ \ ,\ \ g(x)=x^2-3x+2\\\\f(g(x))=\dfrac{2}{x^2-3x+2}\ \ ,\ \ \ \ g(f(x))=\dfrac{4}{x^2}-\dfrac{6}{x}+2$

Answer 2

1. Для f(x) = x^2 и g(x) = 2x - 5:

f(g(x)) = f(2x - 5) = (2x - 5)^2 = 4x^2 - 20x + 25

g(f(x)) = g(x^2) = 2(x^2) - 5 = 2x^2 - 5

2. Для f(x) = x^2 + 1 и g(x) = √(3 - x):

f(g(x)) = f(√(3 - x)) = (√(3 - x))^2 + 1 = (3 - x) + 1 = 4 - x

g(f(x)) = g(x^2 + 1) = √(3 - (x^2 + 1)) = √(2 - x^2)

3. Для f(x) = x^2 - 1 и g(x) = x^2 + 1:

f(g(x)) = f(x^2 + 1) = (x^2 + 1)^2 - 1 = x^4 + 2x^2 + 1 - 1 = x^4 + 2x^2

g(f(x)) = g(x^2 - 1) = (x^2 - 1)^2 + 1 = x^4 - 2x^2 + 1 + 1 = x^4 - 2x^2 + 2

4. Для f(x) = 2/x и g(x) = x^2 - 3x + 2:

f(g(x)) = f(x^2 - 3x + 2) = 2 / (x^2 - 3x + 2)

g(f(x)) = g(2/x) = (2/x)^2 - 3(2/x) + 2 = 4/x^2 - 6/x + 2