Let's solve the problem step by step.

Given:

• $g(x) = x^2 - 1$
• $h(x) = e^x + 1$

i) Suggest a domain for $g(x)$ such that $g^{-1}(x)$ exists.

For $g(x)$ to have an inverse, it must be one-to-one. The given function $g(x) = x^2 - 1$ is a quadratic function, which is not one-to-one over its entire domain ($-\infty, \infty$) because it is symmetric about the $y$-axis.

To make $g(x)$ one-to-one, we must restrict the domain to one side of the vertex. The vertex of $g(x)$ is at $x = 0$, so we can restrict the domain to either:

1. $x \geq 0$ (right side of the vertex), or
2. $x \leq 0$ (left side of the vertex).

Let us choose $x \geq 0$ as the restricted domain. This ensures $g(x)$ is one-to-one.

ii) Given the domain of $h$ is $(-\infty, \infty)$, find an expression for $h^{-1}(x)$ and sketch the graphs of $h(x)$ and $h^{-1}(x)$.

Step 1: Find $h^{-1}(x)$

We are given: $h(x) = e^x + 1$

To find the inverse $h^{-1}(x)$, solve for $x$ in terms of $y$: $y = e^x + 1$ $y - 1 = e^x$ $x = \ln(y - 1)$

Thus: $h^{-1}(x) = \ln(x - 1)$

Step 2: Characteristics of the graphs

1. $h(x)$:

   • Domain: $(-\infty, \infty)$
   • Range: $(1, \infty)$
   • Asymptote: Horizontal asymptote at $y = 1$ as $x \to -\infty$
   • Intercepts: $h(0) = e^0 + 1 = 2$

2. $h^{-1}(x)$:

   • Domain: $(1, \infty)$
   • Range: $(-\infty, \infty)$
   • Asymptote: Vertical asymptote at $x = 1$ as $y \to -\infty$
   • Intercepts: $h^{-1}(2) = \ln(2 - 1) = \ln(1) = 0$

Step 3: Sketch

• Plot $h(x) = e^x + 1$, noting the asymptote $y = 1$, and passing through the point $(0, 2)$.
• Plot $h^{-1}(x) = \ln(x - 1)$, noting the asymptote $x = 1$, and passing through $(2, 0)$.
• Reflect $h(x)$ across the line $y = x$ to get $h^{-1}(x)$.

iii) Determine an expression for $g(h(x))$ in its simplest form.

We are given: $g(x) = x^2 - 1, \quad h(x) = e^x + 1$

Substitute $h(x)$ into $g(x)$: $g(h(x)) = g(e^x + 1)$ $g(h(x)) = (e^x + 1)^2 - 1$

Expand $(e^x + 1)^2$: $g(h(x)) = e^{2x} + 2e^x + 1 - 1$ $g(h(x)) = e^{2x} + 2e^x$

Thus: $g(h(x)) = e^{2x} + 2e^x$

Final Answer:

1. (i): Domain of $g(x)$ for $g^{-1}(x)$ to exist: $x \geq 0$ (or $x \leq 0$).
2. (ii):
   • Expression for $h^{-1}(x): \ln(x - 1)$.
   • Graph characteristics outlined above.
3. (iii): Simplified expression for $g(h(x))$: $g(h(x)) = e^{2x} + 2e^x$.

Would you like me to provide further clarifications or help with the sketch?

Here are 5 additional questions to extend this topic:

1. What is the significance of restricting the domain when finding an inverse?
2. How do horizontal and vertical asymptotes relate to the original and inverse functions?
3. Can $h^{-1}(x)$ exist if the domain of $h(x)$ is altered?
4. How do you confirm that $h^{-1}(h(x)) = x$ holds for all $x$ in the domain?
5. What are the practical applications of composite functions like $g(h(x))$?

Tip: Always verify the domain and range of a function and its inverse to ensure consistency.