Question

Which of the following is an even function g(x) = (x – 1)2 + 1g(x) = 2x2 + 1g(x) = 4x + 2g(x) = 2x

Answer 1

An even function can be reflected about the y axis and map onto itself
example: y=x^2

an odd function can be reflected about the origin and map onto itself
example: y=x^3


a simple test is the following

if f(x) is even then f(-x)=f(x)
if f(x) is odd then f(-x)=-f(x)


so

even function
subsitute -x for each and see if we get the same function
remember to fully expand these

g(x)=(x-1)^2+1=x^2-2x+1+1=x^2-2x+2 is the original one

g(x)=(x-1)^2+1
g(-x)=(-x-1)^2+1
g(-x)=(1)(x+1)^2+1
g(-x)=x^2+2x+1+1
g(-x)=x^2+2x+2
not same because the original has -2x
not even


g(x)=2x^2+1
g(-x)=2(-x)^2+1
g(-x)=2x^2+1
same, it's even

g(x)=4x+2
g(-x)=4(-x)+2
g(-x)=-4x+2
not the same, not even

g(x)=2x
g(-x)=2(-x)
g(-x)=-2x
not same, not even



g(x)=2x²+1 is the even function

Answer 2

Answer:

2nd function is even function.

Step-by-step explanation:

Given Functions are

$1.\: g(x)=(x-1)^2+1\\2.\:g(x)=2.x^2+1\\3.\:g(x)=4.x+2\\4.\:g(x)=2.x$

Even Functions are functions whose value didnt change if we replace the variable with negative of the variable.i.e., f(x) = f(-x)

here,

1st function

$g(x)=(x-1)^2+1\implies g(x)= x^2-2.x+1+1\implies g(x)=x^2-2.x+2$

replacing x by -x we get

$g(-x)=(-x)^2-2.(-x)+2\\g(-x)=x^2+2.x+2\\\implies g(-x)\neq g(x)$

∴ It is not an even function.

2nd function

$g(x)=2.x^2+1$

replacing x by -x we get,

$g(-x)=2.(-x)^2+1\\g(-x)=2.x^2+1\\\implies g(-x)=g(x)$

∴It is an even function.

3rd function

$g(x)=4.x+2$

replacing x by -x we get,

$g(-x)=4.(-x)+2\\g(-x)=-4.x+2\\\implies g(-x)\neq g(x)$

∴It is not an even function

4th function

$g(x)=2.x$

replacing x by -x we get,

$g(-x)=2.(-x)\\g(-x)=-2.x\\\implies g(-x)\neq g(x)$

∴It is not an even function

Therefore, only 2nd function is even function.