which of the following is an even function?\n$g(x)=(x - 1)^2+1$\n$g(x)=2x^2 + 1$\n$g(x)=4x + 2$\n$g(x)=2x$

Explanation:

Step1: Recall the definition of an even function

A function $g(x)$ is even if $g(-x)=g(x)$ for all $x$ in the domain of $g$.

Step2: Test $g(x)=(x - 1)^2+1$

$g(-x)=(-x - 1)^2+1=(x + 1)^2+1=x^{2}+2x + 1+1=x^{2}+2x+2$, and $g(x)=x^{2}-2x + 1+1=x^{2}-2x+2$. Since $g(-x)\neq g(x)$, it is not an even function.

Step3: Test $g(x)=2x^{2}+1$

$g(-x)=2(-x)^{2}+1=2x^{2}+1$. Since $g(-x)=g(x)$, it is an even function.

Step4: Test $g(x)=4x + 2$

$g(-x)=4(-x)+2=-4x + 2$. Since $g(-x)\neq g(x)$, it is not an even function.

Step5: Test $g(x)=2x$

$g(-x)=2(-x)=-2x$. Since $g(-x)\neq g(x)$, it is not an even function.

Answer:

$g(x)=2x^{2}+1$