Question

The function g has an inverse. The function g − 1 determines the number of folds needed to give the folded paper a thickness of t mm. Write a function formula for g − 1 .

Answer 1

To find the inverse function g⁻¹, which calculates the number of folds needed to reach a certain thickness t, you would use the logarithm base 2. The inverse function g⁻¹ is given by g⁻¹(t) = log₂(t).

Step-by-step explanation:

The question asks to write a function formula for g⁻¹, which is the inverse of the function g. The function g⁻¹ determines the number of folds needed to give the folded paper a thickness of t millimeters. To find the inverse, one would typically reverse the operations performed by the original function g.

The typical paper folding problem is that each fold doubles the thickness of the paper. So, if g(n) represents the thickness of the paper after n folds, the thickness is 2n times the original thickness. Hence, the inverse g⁻¹(t) would solve for n given a thickness t, such that t = 2n. This is done by taking the logarithm base 2 of t, which gives g⁻¹(t) = log₂(t).

Answer 2

`\mathbf{g^(-1) (t) = log _2 \ 20 \ t}`

Step-by-step explanation:

Here is the full question

A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times. We'll assume we can make "perfect folds," where each fold makes the folded paper exactly twice as thick as before - and we can make as many folds as we want.

Write a function g that determines the thickness of the folded paper (in mm) in terms of the number folds made, n. (Notice that g(0) 0.05,)

`g(n )= (05)2^n`

The function g has an inverse. The function g⁻¹ determines the number of folds needed to give the folded paper a thickness of t mm. Write a function formula for g⁻¹).

SOLUTION:

If we represent g(n) with t;

Then

`\mathbf{t = (0.05)2^n} \\ \\ \mathbf{(t)/(0.05) = 2^n} \\ \\ \mathbf{\frac {100 \ t }{ 5} = 2^n} \\ \\ \mathbf{20 t = 2^n}`

Taking logarithm of both sides; we have :

`\mathbf{log (20 t) = n log 2} \ \ \ \mathbf{(since \ , log \ a^b = b \ log \ a) } \\ \\ \mathbf{n = (log \ 20 t )/(log \ 2 )} \\ \\ \mathbf{g^(-1) (t) = (log \ 20 t )/(log \ 2 )} \\ \\ \mathbf{g^(-1) (t) = log _2 \ 20 \ t}`