Identify odd functions from equations Precalculus

A function is odd 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, 3-x^2\neq 0,
that is to say x\neq \sqrt{3} and x\neq -\sqrt{3}.
Therefore, the domain of f is \mathcal{D}=\mathbb{R}-\left\{-\sqrt{3},\sqrt{3}\right\}, which is symmetric about 0.
Let x be a real number in \mathcal{D}.

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

We also have:

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

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

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

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

Therefore:

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

A function is odd 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 is symmetric about 0.
Let x be a real number.

f\left(x\right) = x^4 - x^3 + 7

We also have:

f\left(-x\right) = \left(-x\right)^4 + \left(-x\right)^3 + 7

f\left(-x\right) = x^4 - \left(-x^3\right) + 7

f\left(-x\right) = x^4 + x^3 + 7

Note that:

-f\left(x\right) = -\left(x^4\right) + x^3 + 7

Therefore:

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

A function is odd 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^4-x^2\geq 0 and x\neq 0.
x^4-x^2\geq 0 if, and only if, x^2\left(x^2-1\right)\geq 0,
that is to say x\leq -1 or x\geq 1.
Therefore, the domain of f is \mathcal{D}=(-\infty,-1]\cup[1,\infty), which is symmetric about 0.
Let x be a real number in \mathcal{D}.

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

We also have:

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

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

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

Note that:

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

Therefore:

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

A function is odd 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^4\neq 0,
that is to say x\neq 0.
Therefore, the domain of f is \mathcal{D}=\mathbb{R}-\{0\}, which is symmetric about 0.
Let x be a real number in \mathcal{D}.

f\left(x\right) = \dfrac{\sqrt[3]{x}}{x^4}

We also have:

f\left(-x\right) = \dfrac{\sqrt[3]{-x}}{\left(-x\right)^4}

f\left(-x\right) = \dfrac{-\sqrt[3]{x}}{x^4}

f\left(-x\right) = -\dfrac{\sqrt[3]{x}}{x^4}

Note that:

-f\left(x\right) = -\dfrac{\sqrt[3]{x}}{x^4}

Therefore:

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

A function is odd 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^5\neq 0,
that is to say x\neq 0.
Therefore, the domain of f is \mathcal{D}=\mathbb{R}-\{0\}, which is symmetric about 0.
Let x be a real number in \mathcal{D}.

f\left(x\right) = \dfrac{\sqrt[3]{x}}{x^5}

We also have:

f\left(-x\right) = \dfrac{\sqrt[3]{-x}}{\left(-x\right)^5}

f\left(-x\right) = \dfrac{-\sqrt[3]{x}}{-\left(x^5\right)}

f\left(-x\right) = \dfrac{\sqrt[3]{x}}{x^5}

Note that:

-f\left(x\right) = -\dfrac{\sqrt[3]{x}}{x^5}

Therefore:

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

A function is odd 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^3-x\neq 0,
that is to say x\left(x^2-1\right)\neq 0.
Therefore, the domain of f is \mathcal{D}=\mathbb{R}-\{-1{,}0{,}1\}, which is symmetric about 0.
Let x be a real number in \mathcal{D}.

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

We also have:

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

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

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

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

Note that:

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

Therefore:

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

A function is odd 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 is symmetric about 0.
Let x be a real number.

f\left(x\right) = x^5 + 4x^3 - x

We also have:

f\left(-x\right) = \left(-x\right)^5 + 4\left(-x\right)^3 - \left(-x\right)

f\left(-x\right) = -\left(x^5\right) - 4\left(x^3\right) + x

f\left(-x\right) = -\left(x^5 + 4x^3 - x\right)

Note that:

-f\left(x\right) = -\left(x^5 + 4x^3 - x\right)

Therefore:

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