Determine if Odd, Even, or Neither square root of 6x^2+1

$\sqrt{6 x^{2} + 1}$

Step 1

Write $\sqrt{6 x^{2} + 1}$ as a function.

$f (x) = \sqrt{6 x^{2} + 1}$

Step 2

Find $f (- x)$ .

Find $f (- x)$ by substituting $- x$ for all occurrence of $x$ in $f (x)$ .

$f (- x) = \sqrt{6 {(- x)}^{2} + 1}$

Apply the product rule to $- x$ .

$f (- x) = \sqrt{6 ({(- 1)}^{2} x^{2}) + 1}$

Raise $- 1$ to the power of $2$ .

$f (- x) = \sqrt{6 (1 x^{2}) + 1}$

Multiply $x^{2}$ by $1$ .

$f (- x) = \sqrt{6 x^{2} + 1}$

Step 3

A function is even if $f (- x) = f (x)$ .

Check if $f (- x) = f (x)$ .

Since $\sqrt{6 x^{2} + 1} = \sqrt{6 x^{2} + 1}$ , the function is even.

The function is even

Step 4