Question

If f(x) = √2x + 3 and g(x) = x^2, for what value(s) of x does f(g(x)) = g(f(x))? (approximate when needed). Please give an explanation with your answer!​

Answer 1

Answer:

$\sf\large\boxed{x=\sqrt6-3}$

Step-by-step explanation:

$\sf\:Domain:\\2x+3\geq0\to x\geq-1.5$

$\sf\:f(x)=\sqrt{2x+3},\ \sf\:g(x)=x^2\\\\f(g(x))-\text{substitute x = g(x) in}\ f(x):\\\\f(g(x))=f(x^2)=\sqrt{2x^2+3}\\\\g(f(x))-\text{substitute x = f(x) in}\ g(x):\\\\g(f(x))=g(\sqrt{2x+3})=(\sqrt{2x+3})^2=2x+3\\\\f(g(x))=g(f(x))\iff\sqrt{2x^2+3}=2x+3\qquad\text{square of both sides}\\\\(\sqrt{2x^2+3})^2=(2x+3)^2}\qquad\text{use}\ (\sqrt{a})^2=a\ \text{and}\ (a+b)^2=a^2+2ab+b^2\\\\2x^2+3=(2x)^2+2(2x)(3)+3^2\\\\2x^2+3=4x^2+12x+9\qquad\text{subtract}\ 2x^2\ \text{and 3 from both sides}$

$0=2x^2+12x+6\qquad\text{divide both sides by 2}\\\\x^2+6x+3=0\qquad\text{add 6 to both sides}\\\\ \sf\:x^2+6x+9=6\\\\ \sf\:x^2+2(x)(3)+3^2=6\qquad\text{use}\ \sf\:(a+b)^2=a^2+2ab+b^2\\\\ \sf\:(x+3)^2=6\iff x+3=\pm\sqrt6\qquad\text{subtract 3 from both sides}\\\\x=-3-\sqrt6\notin D\ \vee\ x=-3+\sqrt6\in D$