Let's break down each part of the problem step by step. We are given two functions:

$f(x) = 2x^2 \quad \text{and} \quad g(x) = 3x + 1$

1. $f(x) + g(x)$

To find $f(x) + g(x)$, simply add the two functions together:

$f(x) + g(x) = 2x^2 + (3x + 1)$

Combine the like terms:

$f(x) + g(x) = 2x^2 + 3x + 1$

2. $f(x) - g(x)$

Now, subtract $g(x)$ from $f(x)$:

$f(x) - g(x) = 2x^2 - (3x + 1)$

Distribute the minus sign:

$f(x) - g(x) = 2x^2 - 3x - 1$

3. $f[g(x)]$

To find $f[g(x)]$, we need to substitute $g(x)$ into $f(x)$. Since $f(x) = 2x^2$, we substitute $g(x) = 3x + 1$ for $x$ in $f(x)$:

$f[g(x)] = 2(3x + 1)^2$

Now expand $(3x + 1)^2$:

$(3x + 1)^2 = 9x^2 + 6x + 1$

So:

$f[g(x)] = 2(9x^2 + 6x + 1)$

Distribute the 2:

$f[g(x)] = 18x^2 + 12x + 2$

4. $f(x)^2$

Now, square $f(x)$:

$f(x)^2 = (2x^2)^2$

Square the terms:

$f(x)^2 = 4x^4$

5. $f(x)^3 - g(x) + g(x)^2$

This expression involves a combination of powers of $f(x)$ and $g(x)$. First, calculate each part:

$f(x)^3 = (2x^2)^3 = 8x^6$ $g(x) = 3x + 1$ $g(x)^2 = (3x + 1)^2 = 9x^2 + 6x + 1$

Now combine everything:

$f(x)^3 - g(x) + g(x)^2 = 8x^6 - (3x + 1) + (9x^2 + 6x + 1)$

Simplify the expression:

$f(x)^3 - g(x) + g(x)^2 = 8x^6 + 9x^2 + 3x$

Final Answers:

1. $f(x) + g(x) = 2x^2 + 3x + 1$
2. $f(x) - g(x) = 2x^2 - 3x - 1$
3. $f[g(x)] = 18x^2 + 12x + 2$
4. $f(x)^2 = 4x^4$
5. $f(x)^3 - g(x) + g(x)^2 = 8x^6 + 9x^2 + 3x$

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Would you like more details on any of these steps? Let me know if you have any questions!

Here are some related questions:

1. How would you evaluate $f(x) + g(x)$ at a specific value of $x$?
2. What happens if you substitute different functions for $f(x)$ and $g(x)$ in these expressions?
3. How do function compositions like $f[g(x)]$ relate to real-world problems?
4. Can you extend these operations to more than two functions, say $f(x)$, $g(x)$, and $h(x)$?
5. How would the result change if we used a more complex function, like $f(x) = x^3$?

Tip: When working with function operations, always be mindful of the order of operations (like when distributing or composing functions) to avoid mistakes!