Identify even functions from equations Precalculus

A function is even if and only if:

The domain of f, \mathcal{D}, is symmetric about 0 and f(-x)=f(x) for all x in \mathcal{D}.

Here, we have:

f(x) exists if, and only if, 1+x\neq 0,
that is to say x\neq -1.

Therefore, the domain of f is \mathcal{D}=\mathbb{R}-\{-1\}.
And \mathcal{D} is not symmetric about 0.

A function is even if and only if:

The domain of f, \mathcal{D}, is symmetric about 0 and f(-x)=f(x) for all x in \mathcal{D}.

Here, we have:

As x^4\geq 0, x^2\geq 0, we get x^4+x^2+3>0.
Therefore, the domain of f is \mathbb{R},
which is symmetric about 0.
Let x be a real number.

f \left(x\right)=\dfrac{x^2 + 4}{x^4 + x^2 + 3}

And:

f\left(-x\right) = \dfrac{\left(-x\right)^2 + 4}{\left(-x\right)^4 + \left(-x\right)^2 + 3}

f\left(-x\right) = \dfrac{x^2 + 4}{x^4 + x^2 + 3}

So:

f\left(-x\right) = f\left(x\right)

A function is even if and only if:

The domain of f, \mathcal{D}, is symmetric about 0 and f(-x)=f(x) for all x in \mathcal{D}.

Here, we have:

f(x) exists if, and only if, x^2\neq 0, that is to say x\neq 0.
Therefore, the domain of f is \mathbb{R}-\{0\}, which symmetric about 0.
Let x be a real number non equal to 0.

f \left(x\right)= \dfrac{\left(x+3\right)\left(x-2\right)}{x^2}

And:

f\left(-x\right) = \dfrac{\left(-x+3\right)\left(-x-2\right)}{\left(-x\right)^2}

f\left(-x\right) = \dfrac{\left[-\left(x-3\right)\right]\left[-\left(x + 2\right)\right]}{x^2}

f\left(-x\right) = \dfrac{\left(x-3\right)\left(x + 2\right)}{x^2}

So:

f\left(-x\right) \neq f\left(x\right)

A function is even if and only if:

The domain of f, \mathcal{D}, is symmetric about 0 and f(-x)=f(x) for all x in \mathcal{D}.

Here, we have:

The domain of f is \mathbb{R}, which symmetric about 0.
Let x be a real number.

f \left(x\right)= \left(x^4-1\right)\left(x-2\right)^2

And:

f\left(-x\right) = \left(\left(-x\right)^4-1\right)\left(-x-2\right)^2

f\left(-x\right) = \left(x^4-1\right)\left(-1\right)^2\left(x+2\right)^2

f\left(-x\right) = \left(x^4-1\right)\left(x+2\right)^2

Therefore:

f\left(-x\right) \neq f\left(x\right)

A function is even if and only if:

The domain of f, \mathcal{D}, is symmetric about 0 and f(-x)=f(x) for all x in \mathcal{D}.

Here, we have:

The domain of f is \mathbb{R}, which symmetric about 0.
Let x be a real number.

f \left(x\right)= x^4 - x^2 - 2

And:

f\left(-x\right) = \left(-x\right)^4 - \left(-x\right)^2 - 2

f\left(-x\right) = x^4 - x^2 - 2

So:

f\left(-x\right) = f\left(x\right)

A function is even if and only if:

The domain of f, \mathcal{D}, is symmetric about 0 and f(-x)=f(x) for all x in \mathcal{D}.

Here, we have:

As x^4\geq 0, we get x^4+17>0.
Therefore, the domain of f is \mathbb{R},
which is symmetric about 0.
Let x be a real number.

f \left(x\right)= \dfrac{x^6 + 4x^2 - 2}{x^4 + 17}

And:

f\left(-x\right) = \dfrac{\left(-x\right)^6 + 4\left(-x\right)^2 - 2}{\left(-x\right)^4 + 17}

f\left(-x\right) = \dfrac{x^6 + 4x^2 - 2}{x^4 + 17}

So:

f\left(-x\right) = f\left(x\right)

A function is even if and only if:

The domain of f, \mathcal{D}, is symmetric about 0 and f(-x)=f(x) for all x in \mathcal{D}.

Here, we have:

As x^4\geq 0, x^2\geq 0, we get x^4+x^2\geq 0.
And x^4+x^2=0 if, and only if, x^2\left(x^2+1\right)=0,
that is to say x=0
Therefore, the domain of f is \mathbb{R}-\{0\},
which is symmetric about 0.
Let x be a real number non equal to 0.

f \left(x\right)= \dfrac{x^6 + 4x^2 - 2x}{x^4 + x^2}

And:

f\left(-x\right) = \dfrac{\left(-x\right)^6 + 4\left(-x\right)^2 - 2\left(-x\right)}{\left(-x\right)^4 + \left(-x\right)^2}

f\left(-x\right) = \dfrac{x^6 + 4x^2 + 2x}{x^4 + x^2}

So:

f\left(-x\right) \neq f\left(x\right)