To prove algebraically that the function $f(x) = 3|x| + x^2$ is an even function, we need to demonstrate that:

$f(-x) = f(x)$

Step-by-step proof:

1. Write the function: $f(x) = 3|x| + x^2$

2. Substitute $-x$ for $x$ into the function: $f(-x) = 3|-x| + (-x)^2$

3. Simplify:

   • The absolute value of $-x$ is the same as $|x|$, since $|-x| = |x|$.
   • $(-x)^2$ is the same as $x^2$, because squaring a negative number results in a positive value.

   So, we get: $f(-x) = 3|x| + x^2$

4. Compare $f(-x)$ with $f(x)$: $f(-x) = 3|x| + x^2 \quad \text{and} \quad f(x) = 3|x| + x^2$ Since $f(-x) = f(x)$, the function is even.

Thus, the function $f(x) = 3|x| + x^2$ is indeed an even function.

Would you like more details or have any questions?

5 related questions:

1. What is the difference between even and odd functions?
2. How do absolute values affect the parity of functions?
3. Can a function be neither even nor odd? If so, provide an example.
4. What is the geometric interpretation of even functions?
5. How can we test whether a polynomial is even or odd?

Tip: To quickly check if a function is even, you only need to substitute $-x$ and see if you get the original function back.