#1**+10 **

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

#1**+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