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3 years ago

Find a formula for the inverse of the function f(x)=(1+e^x)/(1-e^x)

1 answer:

4 0

Let y = f(x)

y = (1 + e^x)/(1 - e^x)

Exchange places with x and y.

x = (1 + e^y)/(1 - e^y)

Now solve for y.

y = log[(x + 1)/(x - 1)]

Replace y with the inverse notation.

f^(-1) x = log[(x + 1)/(x - 1)]

Done!

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Which set of points represents equivalent ratios and would follow the same pattern on a graph?

Paladinen [302]

(2,4), (4,8), (3,6), (5,10) because they’re all ratios of 1:2

8 0

3 years ago

Give this problem a try and try to solve this

tia_tia [17]

Answer:

No solution

Step-by-step explanation:

Given equation is,

$\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1+x^\frac{1}{2}}-\frac{(4+x)^\frac{1}{2}}{(1-x)^\frac{1}{2}}=0$

$\frac{x^{\frac{1}{2}}+x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1+x^\frac{1}{2}}=\frac{(4+x)^\frac{1}{2}}{(1-x)^\frac{1}{2}}$

$\frac{(x+1)}{\sqrt{x}(1-x)}+\frac{(\sqrt{x}-1)}{\sqrt{x}(1+\sqrt{x})}=(\frac{4+x}{1-x})^{\frac{1}{2}}$

$\frac{(\sqrt{x}+1)(x+1)+(\sqrt{x}-1)(1-x)}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^{\frac{1}{2}}$

$\frac{x\sqrt{x}+x+\sqrt{x}+1+\sqrt{x}-1-x\sqrt{x}+x}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}$

$\frac{2x+2\sqrt{x}}{\sqrt{x}(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}$

$\frac{2(\sqrt{x}+1)}{(1-x)(1+\sqrt{x})}=(\frac{4+x}{1-x})^\frac{1}{2}$

$\frac{2}{1-x}=(\frac{4+x}{1-x})^\frac{1}{2}$ if x ≠ ±1

$(\frac{2}{1-x})^2=\frac{4+x}{1-x}$ [Squaring on both the sides of the equation]

$\frac{4}{(1-x)}=(4+x)$

4 = (1 - x)(4 + x)

4 = 4 - 4x + x - x²

0 = -3x - x²

x² + 3x = 0

x(x + 3) = 0

x = 0, -3

But both the solutions x = 0 and x = -3 are extraneous solutions, given equation has no solution.

5 0

3 years ago

Read 2 more answers

Which are equivalent expressions?

ollegr [7]

Answer:

Check Below.

Step-by-step explanation:

x - y = (6 - 5)(y - x) → False

(x - y) (6 - 5) = 1 → False

x- y = (1)(x - y) → True

x - y = 1 + (x - y) → No solution, False

x - y = (6 - 5)(x - y) → True

6 0

2 years ago

Which of the following are compound events in which the results of one event affect the likelihood of the other event?

Rama09 [41]

Answer:

Dependent Events

Step-by-step explanation:

The answer is dependent events because a dependent event relies on the outcome of another event. Remember dependent depends on another event.

3 0

3 years ago

True or false??????

Katyanochek1 [597]

Answer:

true

Step-by-step explanation:

$\sqrt{25} = ±5$

6 0

3 years ago