Question

G(x) = 2x^2 and h(x) = √x^2+1 .What is (goh)^-1 and is it a function?

Answer 1

Answer:

Step-by-step explanation:

Hello,

$(goh)(x)=g(h(x))=2\left( \sqrt{x^2+1}\right)^2=2(x^2+1)\\\\x=(goh)((goh)^{-1}(x))=2((goh)^{-1}(x)^2+1)\\ \\\left((goh)^{-1}(x)\right)^2=\dfrac{x}{2}-1=\dfrac{x-2}{2}\\ \\(goh)^{-1}(x)=\sqrt{\dfrac{x-2}{2}}$

And this is a function defined for x-2 $\geq$ 0, meaning x $\geq$ 2

Thanks