Determine if Odd, Even, or Neither f(x) = square root of 5x^2+1

$f (x) = \sqrt{5 x^{2} + 1}$

Step 1

Find $f (- x)$ .

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

$f (- x) = \sqrt{5 {(- x)}^{2} + 1}$

Apply the product rule to $- x$ .

$f (- x) = \sqrt{5 ({(- 1)}^{2} x^{2}) + 1}$

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

$f (- x) = \sqrt{5 (1 x^{2}) + 1}$

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

$f (- x) = \sqrt{5 x^{2} + 1}$

Step 2

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

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

Since $\sqrt{5 x^{2} + 1} = \sqrt{5 x^{2} + 1}$ , the function is even.

The function is even

Step 3