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

`\large\boxed{x=\sqrt6-3}`

Explanation:

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

`f(x)=√(2x+3),\ g(x)=x^2\\\\f(g(x))-\text{substitute x = g(x) in}\ f(x):\\\\f(g(x))=f(x^2)=√(2x^2+3)\\\\g(f(x))-\text{substitute x = f(x) in}\ g(x):\\\\g(f(x))=g(√(2x+3))=(√(2x+3))^2=2x+3\\\\f(g(x))=g(f(x))\iff√(2x^2+3)=2x+3\qquad\text{square of both sides}\\\\(√(2x^2+3))^2=(2x+3)^2}\qquad\text{use}\ (√(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}\\\\x^2+6x+9=6\\\\x^2+2(x)(3)+3^2=6\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+3)^2=6\iff x+3=\pm\sqrt6\qquad\text{subtract 3 from both sides}\\\\x=-3-\sqrt6\\otin D\ \vee\ x=-3+\sqrt6\in D`