To identify a function \(\displaystyle f(x)\) as even odd, or neither, determine \(\displaystyle f(-x)\) by replacing \(\displaystyle x\) with \(\displaystyle -x\), then simplifying. If \(\displaystyle f(-x) = f(x)\), the function is even; if \(\displaystyle f(-x) = - f(x)\) is odd.

\(\displaystyle f(x) = 3x^{3}+ 4x\),

so

\(\displaystyle f(-x) = 3(-x)^{3}+ 4(-x)\)

By the Power of a Product Property,

\(\displaystyle f(-x) = 3(-1)^{3} x^{3}+ 4(-x)\)

\(\displaystyle f(-x) = 3(-1) x^{3}+ 4(-x)\)

\(\displaystyle f(-x) = -3x^{3}-4x\)

\(\displaystyle f(-x) = -3x^{3}-4x = -(3x^{3}+ 4x) = -f(x)\),

so \(\displaystyle f(x)\) is an odd function

To identify a function \(\displaystyle f(x)\) as even odd, or neither, determine \(\displaystyle f(-x)\) by replacing \(\displaystyle x\) with \(\displaystyle -x\), then simplifying. If \(\displaystyle f(-x) = f(x)\), the function is even; if \(\displaystyle f(-x) = - f(x)\) is odd.

\(\displaystyle f(x) = x^{7}+ x^{6}- x\)

so

\(\displaystyle f(-x) = (-x) ^{7}+ (-x) ^{6}- (-x)\)

By the Power of a Product Property,

\(\displaystyle f(-x) =(-1)^{7} x ^{7}+(-1)^{6} x ^{6}- (-x)\)

\(\displaystyle f(-x) =(-1) x ^{7}+1 x ^{6}+x\)

\(\displaystyle f(-x) =-x ^{7}+ x ^{6}+x\)

\(\displaystyle f(-x) \ne f(x)\), so \(\displaystyle f(x)\) is not an even function.

\(\displaystyle - f(x) = - (x^{7}+ x^{6}- x) = - x^{7}- x^{6} + x\),

\(\displaystyle f(-x) \ne -f(x)\), so \(\displaystyle f(x)\) is not an odd function.

To identify a function \(\displaystyle f(x)\) as even, odd, or neither, determine \(\displaystyle f(-x)\) by replacing \(\displaystyle x\) with \(\displaystyle -x\), then simplifying. If \(\displaystyle f(-x) = f(x)\), the function is even; if \(\displaystyle f(-x) = - f(x)\) is odd.

\(\displaystyle f(x) = \log (x^{2})\)

\(\displaystyle f(-x) = \log[ (-x)^{2} ] = \log\left ( x^{2} \right ) = f(x)\)

\(\displaystyle f(-x) = f(x)\), so \(\displaystyle f(x)\) is an even function.

To identify a function \(\displaystyle f(x)\) as even, odd, or neither, determine \(\displaystyle f(-x)\) by replacing \(\displaystyle x\) with \(\displaystyle -x\), then simplifying. If \(\displaystyle f(-x) = f(x)\), the function is even; if \(\displaystyle f(-x) = - f(x)\) is odd.

\(\displaystyle f(x) = 4 ^{x}- \frac{1}{4^{x}}\)

\(\displaystyle f(-x) = 4 ^{-x}- \frac{1}{4^{-x}}\)

\(\displaystyle f(-x) = \frac{1}{4^{x}}- 4 ^{x}\)

\(\displaystyle -f(x) =- \left (4 ^{x}- \frac{1}{4^{x}} \right ) = -4 ^{x}+ \frac{1}{4^{x}} = \frac{1}{4^{x}} -4 ^{x} = f(-x)\)

Since \(\displaystyle f(-x) = - f(x)\), \(\displaystyle f(x)\) is an odd function.

A function is odd if and only if, for every \(\displaystyle c\) in its domain, \(\displaystyle f(-c) = -f(c)\); it is even if and only if, for every \(\displaystyle c\) in its domain, \(\displaystyle f(-c) = f(c)\). We can see that

\(\displaystyle f(-3) = 5 = -(-5) = - f(3)\)

\(\displaystyle f(-2) = 7 = -(-7) = - f(2)\)

\(\displaystyle f(-1) = 2 = -(-2) = - f(1)\)

\(\displaystyle f(-0) = 0 = -0 = -f(0)\)

It follows that \(\displaystyle f(x)\) is an odd function.

A function is odd if and only if, for every \(\displaystyle c\) in its domain, \(\displaystyle f(-c) = -f(c)\); it is even if and only if, for every \(\displaystyle c\) in its domain, \(\displaystyle f(-c) = f(c)\). We can see that 

\(\displaystyle f(-3) = 9 = -(-9) = - f(3)\);

the function cannot be even. This does allow for the function to be odd. However, if \(\displaystyle f(x)\) is odd, then, by definition, 

\(\displaystyle f(-0) = -f(0)\), or

\(\displaystyle f(0) = -f(0)\)

and \(\displaystyle f(0)\) is equal to its own opposite - the only such number is 0, so

\(\displaystyle f(0) = 0\).

This is not the case - \(\displaystyle f(0) = 1\) - so the function is not odd either.

A function \(\displaystyle f(x)\) is odd if and only if, for every \(\displaystyle c\) in its domain, \(\displaystyle f(-c) = -f(c)\); it is even if and only if, for every \(\displaystyle c\) in its domain, \(\displaystyle f(-c) = f(c)\). We can see that 

\(\displaystyle f(-3) = 3 = f(3)\)

\(\displaystyle f(-2) =9 = f(2)\)

\(\displaystyle f(-1) = -1 = f(1)\)

Of course, 

\(\displaystyle f(0) = f(-0)\).

Therefore, \(\displaystyle f(x)\) is even by definition.

The relation graphed is a function, as it passes the vertical line test - no vertical line can pass through it more than once, as is demonstrated below:

Also, it can be seen to be symmetrical about the origin. Consequently, for each \(\displaystyle c\) in the domain, \(\displaystyle f(-c) = -f(c)\) - the function is odd.

The vertical asymptote(s) of the graph of a rational function such as \(\displaystyle f(x)\) can be found by evaluating the zeroes of the denominator after the rational expression is reduced. The expression is in simplest form, so set the denominator equal to 0 and solve for \(\displaystyle x\):

\(\displaystyle x-6 = 0\)

\(\displaystyle x= 6\)

The graph of \(\displaystyle f(x)\) has the line of the equation \(\displaystyle x= 6\) as its only vertical asymptote.

The vertical asymptote(s) of the graph of a rational function such as \(\displaystyle f(x)\) can be found by evaluating the zeroes of the denominator after the rational expression is reduced. 

First, factor the numerator. It is a quadratic trinomial with lead term \(\displaystyle x^{2}\), so look to "reverse-FOIL" it as

\(\displaystyle x^{2} - 11x+ 30 = (x+a)(x+b)\)

by finding two integers with sum \(\displaystyle -11\) and product 30. By trial and error, these integers can be found to be \(\displaystyle -6\) and \(\displaystyle -5\), so

\(\displaystyle x^{2} - 11x+ 30 = (x-5)(x-6)\)

Therefore, \(\displaystyle f(x)\) can be rewritten as 

\(\displaystyle f(x) = \frac{(x-5)(x-6)}{x-6}\).

Cancelling \(\displaystyle x -6\), this can be seen to be essentially a polynomial function:

\(\displaystyle f(x) =x-5\),

which does not have a vertical asymptote.