Let's analyze and solve the given problem. The functions are:

$f(x) = 5e^x \quad \text{and} \quad g(x) = x^5$

We are tasked with finding the following values:

(a) $f(g(1))$

This means we first calculate $g(1)$, then use that result as the input to $f$.

$g(1) = 1^5 = 1$ Now, substitute into $f$: $f(1) = 5e^1 = 5e$ Thus, $f(g(1)) = 5e$.

(b) $g(f(1))$

First, calculate $f(1)$: $f(1) = 5e^1 = 5e$ Now, use this in $g$: $g(5e) = (5e)^5 = 3125e^5$ Thus, $g(f(1)) = 3125e^5$.

(c) $f(g(x))$

We are asked to find the expression for $f(g(x))$, which is $f(x^5)$: $f(g(x)) = f(x^5) = 5e^{x^5}$ Thus, $f(g(x)) = 5e^{x^5}$.

(d) $g(f(x))$

We are asked to find the expression for $g(f(x))$, which is $g(5e^x)$: $g(f(x)) = (5e^x)^5 = 3125e^{5x}$ Thus, $g(f(x)) = 3125e^{5x}$.

(e) $f(t)g(t)$

This is simply the product of $f(t)$ and $g(t)$: $f(t) = 5e^t, \quad g(t) = t^5$ $f(t)g(t) = (5e^t)(t^5) = 5t^5e^t$ Thus, $f(t)g(t) = 5t^5e^t$.

Final answers:

• (a) $f(g(1)) = 5e$
• (b) $g(f(1)) = 3125e^5$
• (c) $f(g(x)) = 5e^{x^5}$
• (d) $g(f(x)) = 3125e^{5x}$
• (e) $f(t)g(t) = 5t^5e^t$

Would you like more details or have any questions?

Here are five related questions to expand on this:

1. What would be $f(g(2))$ if $g(x) = x^4$ instead?
2. How does the chain rule apply when differentiating $f(g(x))$?
3. What is the inverse of $f(x) = 5e^x$?
4. How would the results change if $g(x)$ were a different polynomial function, such as $g(x) = x^2$?
5. How would you solve for $x$ if $f(x) = g(x)$?

Tip: Understanding the composition of functions $f(g(x))$ or $g(f(x))$ helps in calculus, particularly when dealing with the chain rule for derivatives.