Problem Breakdown

We are tasked with comparing two functions:

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

Let's analyze the problem step by step.

Part a) Find a function $h(x)$ such that solving $h(x) < 0$ will solve the original problem of $f(x) < g(x)$.

We are looking for the values of $x$ where $f(x)$ is less than $g(x)$, i.e.,

$f(x) < g(x)$

Subtract $g(x)$ from both sides to form a new function $h(x)$:

$f(x) - g(x) < 0$

Notice that:

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

So, $f(x) - g(x) = 0$.

This shows that:

$f(x) = g(x)$

for all values of $x$, meaning that $f(x)$ is never less than $g(x)$, and hence no such $x$ exists where $f(x) < g(x)$.

Therefore, there is no solution for $f(x) < g(x)$. In other words, the inequality $f(x) < g(x)$ does not hold for any value of $x$.

Part b) Find the domain of $h(x)$ in interval notation.

Since $h(x) = f(x) - g(x) = 0$ for all $x$, there is no domain restriction for $h(x)$ because it is simply the constant function 0. Therefore, the domain of $h(x)$ is the same as the domain of the original functions $f(x)$ and $g(x)$, which is all real numbers.

Thus, the domain of $h(x)$ is:

$(-\infty, \infty)$

Summary

• a) The function $h(x)$ is $0$ because $f(x) = g(x)$ for all $x$. Therefore, there is no solution to the inequality $f(x) < g(x)$.
• b) The domain of $h(x)$ is $(-\infty, \infty)$.

Would you like a further explanation or have any other questions on this? Here's a related set of questions for further practice:

1. What happens when $f(x) = g(x)$?
2. How can we compare functions if they have similar forms, but slightly different expressions?
3. Can a function like $f(x)$ and $g(x)$ ever be strictly greater than or less than each other? Why or why not?
4. How do exponential terms impact the comparison of functions?
5. If $f(x) \neq g(x)$, how would you approach finding where $f(x) < g(x)$?

Tip: When comparing two functions, always subtract them first to see if the result simplifies the inequality or if it reveals a relationship between the two.