Question

4. Даны функции f(x)=x^ 2 ,g(x)= 1 2x+1 ,h(x)= sqrt 1-x a) Сравните f(h(-1)) и f(g(-1)). b) Составьте ​

Answer 1

$h(x)=\sqrt{1-x}\\\\h(-1)=\sqrt{1-(-1)} =\sqrt{2}\\\\f(x)=x^{2} \\\\f\Big(h(-1)\Big)=f(\sqrt{2})=(\sqrt{2})^{2} =2\\\\\boxed{f\Big(h(-1)\Big)=2}\\\\\\g(x)=\dfrac{1}{2x+1} \\\\g(-1)=\dfrac{1}{2\cdot(-1)+1} =\dfrac{1}{-2+1}=-1\\\\f\Big(g(-1)\Big)=f(-1)=\sqrt{(-1)^{2} } =1\\\\\boxed{f\Big(g(-1)\Big)=1}\\\\\\\boxed{f\Big(h(-1)\Big)>f\Big(g(-1)\Big)}$

$f\Big(h(x)\Big)=\Big(\sqrt{1-x}\Big)^{2} =1-x\\\\g\Big(f(h(x))\Big)=g(1-x)=\dfrac{1}{2(1-x)+1 } =\dfrac{1}{2-2x+1}=\boxed{\frac{1}{3-2x}} \\\\\\g(x)=\dfrac{1}{2x+1} \\\\2x+1=\dfrac{1}{g(x)}\\\\2x=\dfrac{1}{g(x)}-1\\\\2x=\dfrac{1-g(x)}{g(x)}\\\\x=\dfrac{1-g(x)}{2g(x)}$

Обратная функция :

$\boxed{g(x)=\dfrac{1-x}{2x}}$