Here is the solution to all parts of the problem, along with a neatly formatted answer sheet:

Problem Analysis:

We are tasked with working with hyperbolic cosine functions ($\cosh(x)$) to describe catenary curves under certain transformations. Specifically:

1. Translate the $\cosh(x)$ function to pass through given points.
2. Apply further translations to model different chains.
3. Rescale the function uniformly.

Solutions

Part 1: Translation for $f(x)$

We need to find a translated version of $\cosh(x)$ such that the curve passes through the two points $(-1, 0)$ and $(1, 0)$.

The general form for translating the $\cosh(x)$ function is: $f(x) = \cosh(x) - \cosh(c)$ Here, $c$ represents the value that shifts the $\cosh(x)$ curve vertically so that $f(-1) = 0$ and $f(1) = 0$.

Substitute $x = 1$ or $x = -1$ to verify the curve passes through $0$: $f(-1) = \cosh(-1) - \cosh(1) = 0$ $f(1) = \cosh(1) - \cosh(1) = 0$

Thus, the function $f(x)$ is: $\boxed{f(x) = \cosh(x) - \cosh(1)}$

Part 2: Translation for $g(x)$

For this part, the chain is hanging between $(-2, 0)$ and $(2, 0)$. Again, we translate the $\cosh(x)$ function vertically by subtracting $\cosh(c)$, where $c = 2$.

The function becomes: $g(x) = \cosh(x) - \cosh(2)$

Substitute $x = -2$ and $x = 2$ to confirm that $g(x)$ passes through $0$: $g(-2) = \cosh(-2) - \cosh(2) = 0$ $g(2) = \cosh(2) - \cosh(2) = 0$

Thus, the function $g(x)$ is: $\boxed{g(x) = \cosh(x) - \cosh(2)}$

Part 3: Uniform Rescaling to Get $h(x)$

Here, we uniformly compress $g(x)$ vertically and horizontally by a factor of $2$.

1. Horizontal Compression by Factor 2: Replace $x$ with $2x$ in $g(x)$: $g(2x) = \cosh(2x) - \cosh(2)$

2. Vertical Compression by Factor 2: Multiply the entire expression by $\frac{1}{2}$: $h(x) = \frac{1}{2} \left(\cosh(2x) - \cosh(2)\right)$

Thus, the function $h(x)$ is: $\boxed{h(x) = \frac{1}{2} \left(\cosh(2x) - \cosh(2)\right)}$

Final Answer Sheet

1. Function $f(x)$: $f(x) = \cosh(x) - \cosh(1)$

2. Function $g(x)$: $g(x) = \cosh(x) - \cosh(2)$

3. Function $h(x)$: $h(x) = \frac{1}{2} \left(\cosh(2x) - \cosh(2)\right)$

Next Steps:

Would you like a breakdown of the hyperbolic cosine properties, or a detailed graph of these functions? Let me know if you have further questions or need more explanation.

5 Related Questions:

1. What are the key properties of hyperbolic cosine ($\cosh(x)$) that make it suitable for modeling catenary curves?
2. How do transformations affect the shape and position of $\cosh(x)$?
3. What is the significance of the parameter $c$ in translations like $\cosh(x) - \cosh(c)$?
4. How would you model a chain hanging between two points of unequal height?
5. What happens to $h(x)$ if you apply only a vertical or horizontal compression?

Tip:

Remember that the $\cosh(x)$ function is symmetric about $x = 0$, which simplifies transformations and makes it a natural fit for catenary curves!