The user wants me to find the absolute maximum values of three functions, $f(x)$, $g(x)$, and $h(x)$, on the interval $[0, 1]$ and compare them.

The functions are:

• $f(x) = e^{x^2} + e^{-x^2}$
• $g(x) = x e^{x^2} + e^{-x^2}$
• $h(x) = x^2 e^{x^2} + e^{-x^2}$

The absolute maximum of a continuous function on a closed interval occurs either at the endpoints of the interval or at critical points within the interval.

Step 1: Analyze the function f(x)

1. Find the derivative of $f(x)$: $f'(x) = \frac{d}{dx} (e^{x^2} + e^{-x^2}) = e^{x^2}(2x) + e^{-x^2}(-2x) = 2x(e^{x^2} - e^{-x^2})$
2. Find critical points by setting $f'(x)=0$: $2x(e^{x^2} - e^{-x^2}) = 0$. This implies $x=0$ or $e^{x^2} = e^{-x^2}$, which means $x^2 = -x^2$, so $x=0$. The only critical point is at $x=0$, which is an endpoint.
3. Determine the behavior of $f(x)$ on $[0,1]$: For $x \in (0, 1]$, we have $x^2 > 0$, so $-x^2 < x^2$. Since the exponential function $e^t$ is strictly increasing, $e^{-x^2} < e^{x^2}$, which means $e^{x^2} - e^{-x^2} > 0$. Also, $2x > 0$. Therefore, $f'(x) > 0$ for all $x \in (0, 1]$.
4. This means $f(x)$ is a strictly increasing function on the interval $[0, 1]$. The absolute maximum must occur at the right endpoint, $x=1$.
5. Calculate the maximum value, $a$: $a = f(1) = e^{1^2} + e^{-1^2} = e + e^{-1} = e + \frac{1}{e}$

Step 2: Analyze the function h(x)

1. Find the derivative of $h(x)$: $h'(x) = \frac{d}{dx} (x^2 e^{x^2} + e^{-x^2}) = (2x e^{x^2} + x^2 e^{x^2}(2x)) + e^{-x^2}(-2x) = 2x e^{x^2} + 2x^3 e^{x^2} - 2x e^{-x^2}$ $h'(x) = 2x(e^{x^2}(1+x^2) - e^{-x^2})$
2. Find critical points by setting $h'(x)=0$: One solution is $x=0$. For other solutions, we need $e^{x^2}(1+x^2) - e^{-x^2} = 0$, which is equivalent to $(1+x^2) = e^{-2x^2}$. Let $k(x) = (1+x^2) - e^{-2x^2}$. We see that $k(0) = 1-1=0$. Let's check the derivative of $k(x)$: $k'(x) = 2x - e^{-2x^2}(-4x) = 2x(1 + 2e^{-2x^2})$. For $x \in (0, 1]$, $k'(x) > 0$. So $k(x)$ is strictly increasing on $[0,1]$. Since $k(0)=0$, $k(x)>0$ for all $x \in (0,1]$.
3. This means the term $(e^{x^2}(1+x^2) - e^{-x^2})$ is positive for $x \in (0,1]$. Therefore, $h'(x) > 0$ for all $x \in (0,1]$.
4. This means $h(x)$ is also a strictly increasing function on the interval $[0, 1]$. The absolute maximum must occur at the right endpoint, $x=1$.
5. Calculate the maximum value, $c$: $c = h(1) = 1^2 e^{1^2} + e^{-1^2} = e + e^{-1} = e + \frac{1}{e}$

Step 3: Analyze the function g(x) and compare the functions

Instead of analyzing the derivative of $g(x)$, which is more complex, we can compare the functions directly.

1. Compare $f(x)$ and $g(x)$: $f(x) - g(x) = (e^{x^2} + e^{-x^2}) - (x e^{x^2} + e^{-x^2}) = e^{x^2} - x e^{x^2} = (1-x)e^{x^2}$. For $x \in [0, 1]$, $(1-x) \ge 0$ and $e^{x^2} > 0$. So, $f(x) - g(x) \ge 0$, which means $f(x) \ge g(x)$ for all $x \in [0, 1]$.

2. Compare $g(x)$ and $h(x)$: $g(x) - h(x) = (x e^{x^2} + e^{-x^2}) - (x^2 e^{x^2} + e^{-x^2}) = x e^{x^2} - x^2 e^{x^2} = x(1-x)e^{x^2}$. For $x \in [0, 1]$, $x \ge 0$ and $(1-x) \ge 0$, and $e^{x^2} > 0$. So, $g(x) - h(x) \ge 0$, which means $g(x) \ge h(x)$ for all $x \in [0, 1]$.

3. Combining these results, we have the inequality: $f(x) \ge g(x) \ge h(x)$ for all $x \in [0, 1]$.

Step 4: Determine the relationship between a, b, and c

Let $a, b, c$ be the absolute maximum values of $f, g, h$ respectively. From Step 1, $a = \max_{x \in [0,1]} f(x) = f(1) = e + \frac{1}{e}$. From Step 2, $c = \max_{x \in [0,1]} h(x) = h(1) = e + \frac{1}{e}$. So, we already know $a=c$.

Now let's find $b = \max_{x \in [0,1]} g(x)$. From the inequality $f(x) \ge g(x)$, we can say that the maximum of $g(x)$ cannot be greater than the maximum of $f(x)$. So, $b \le a$. Thus, $b \le e + \frac{1}{e}$.

The maximum value of a function must be at least its value at any point in the domain. So, $b$ must be greater than or equal to $g(1)$. $g(1) = 1 \cdot e^{1^2} + e^{-1^2} = e + \frac{1}{e}$. So, $b \ge e + \frac{1}{e}$.

Combining the two inequalities for $b$, we have $b \le e + \frac{1}{e}$ and $b \ge e + \frac{1}{e}$. This implies $b = e + \frac{1}{e}$.

Therefore, we have $a = b = c = e + \frac{1}{e}$.

Conclusion: The absolute maximums are $a = e + 1/e$, $b = e + 1/e$, and $c = e + 1/e$. Hence, $a=b=c$. This corresponds to option D.