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What is the inverse of y=2x^2+3?

gibsonj338  May 16, 2015

Best Answer 

 #1
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y=2x^2 + 3       subtract 3 from both sides

 

y - 3  = 2x^2   divide both sides by 2

 

[y - 3] / 2 =  x^2      take the square root of both sides

 

±√[( y - 3 ) / 2 ]  = x    "swap" x and y

 

±√[( x - 3 ) / 2 ]  = y      for y, write f-1(x)   {signifying the inverse}

 

f-1(x) = ±√[( x - 3 ) / 2 ] 

 

 

Because the original function isn't "one-to-one," it only has an inverse if we restrict its domain.

If we restrict the domain of the original function to (-∞, 0), -√[( x - 3 ) / 2 ]  is the inverse

If we restrict the domain of the original function to (0, ∞ ), √[( x - 3 ) / 2 ] is the inverse

Note that the points (0,3) and (3,0) are ambiguous......they could belong to either restricted function and its respective inverse.

 

Here's a graph......https://www.desmos.com/calculator/bnxeffgjt3

 

Both the resticted domain functions and their respective inverses are symmetric to the line y = x.......as we would expect.......

 

 

CPhill  May 16, 2015
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1+0 Answers

 #1
avatar+78620 
+10
Best Answer

 

 

 

y=2x^2 + 3       subtract 3 from both sides

 

y - 3  = 2x^2   divide both sides by 2

 

[y - 3] / 2 =  x^2      take the square root of both sides

 

±√[( y - 3 ) / 2 ]  = x    "swap" x and y

 

±√[( x - 3 ) / 2 ]  = y      for y, write f-1(x)   {signifying the inverse}

 

f-1(x) = ±√[( x - 3 ) / 2 ] 

 

 

Because the original function isn't "one-to-one," it only has an inverse if we restrict its domain.

If we restrict the domain of the original function to (-∞, 0), -√[( x - 3 ) / 2 ]  is the inverse

If we restrict the domain of the original function to (0, ∞ ), √[( x - 3 ) / 2 ] is the inverse

Note that the points (0,3) and (3,0) are ambiguous......they could belong to either restricted function and its respective inverse.

 

Here's a graph......https://www.desmos.com/calculator/bnxeffgjt3

 

Both the resticted domain functions and their respective inverses are symmetric to the line y = x.......as we would expect.......

 

 

CPhill  May 16, 2015

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