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Thepotemich [5.8K]
3 years ago
12

If y varies inversely as x2 and y = 3 when x = 2, find the equation that connects x and y.

Mathematics
1 answer:
Bond [772]3 years ago
3 0
<h3>Answer:  y = 12/(x^2)</h3>

============================================================

Explanation:

"y varies inversely as x^2" means y = k/(x^2) for some constant k.

Plug in (x,y) = (2,3) and solve for k

y = k/(x^2)

3 = k/(2^2)

3 = k/4

3*4 = k

12 = k

k = 12

The original equation updates to y = 12/(x^2)

As a check, plugging in x = 2 should lead to y = 3

y = 12/(x^2)

y = 12/(2^2)

y = 12/4

y = 3 ... we get the proper y value, so the answer is confirmed.

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Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

h^{-1}(x)=\sqrt{\frac{x+1}{3}} ⇒ answer D

Step-by-step explanation:

* <em>Lets explain how to find the inverse of a function</em>

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = f^{-1}(x)

* <em>Lets use these steps to solve the problems</em>

∵ f(x)=\sqrt{x-3}

∵ f(x) = y

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∴ f^{-1}(x)=g(x)

∴ <u><em>The functions are inverses to each other</em></u>

* <em>Now lets find f(g(x))</em>

- To find f(g(x)) substitute x in f(x) by g(x)

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∴ f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x

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* <em>Lets find the inverse of h(x)</em>

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

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- Take √ for both sides

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∴ We will chose the positive value of the square root

∴ \sqrt{\frac{x+1}{3}}=y

- replace y by h^{-1}(x)

∴ h^{-1}(x)=\sqrt{\frac{x+1}{3}}

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