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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.

Show algebraically that the function f(x) = 3|x| + x^2 is an even function.

Explore how to prove algebraically that the function f(x) = 3|x| + x^2 is an even function. This detailed guide covers the concept of even functions, properties of absolute value, and includes a step-by-step algebraic proof. Ideal for high school students learning advanced algebra.

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