I have been given that f(x) = 3x^2 - 4 and x > 0 ... and I have been asked to find the value of f^-1(x) when x > -4 ...

Let y = 3x^2 - 4, then

Solving this equation:

or

Either of them is wrong.. The answer should be

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- April 20th 2009, 02:21 AMstruckInverse functions problem
I have been given that f(x) = 3x^2 - 4 and x > 0 ... and I have been asked to find the value of f^-1(x) when x > -4 ...

Let y = 3x^2 - 4, then

Solving this equation:

or

Either of them is wrong.. The answer should be - April 20th 2009, 03:43 AMmr fantastic
- April 20th 2009, 06:11 AMstapel
You are given a quadratic whose vertex is at (h, k) = (0, -4). You are told to consider only the right-hand half of the parabola; with this restriction, the function passes the Horizontal Line Test, and is invertible.

From what you know about**graphing quadratics**, you know that the graph of f(x), for x > 0, is the right-hand half of a right-side up parabola, with domain x > 0 and range y > -4. The inverse will then be the upper half of a rightward-opening sideways parabola, with domain x > -4 and range y > 0.

To learn the general process for finding an inverse, try**here**. Then:

You have:

. . . . .

. . . . .

To solve for "x=", apply**the Quadratic Formula**:

. . . . .

. . . . .

. . . . . . . .

Since you know you need the range to be y > 0, then you know to take only the "plus" part of the "plus, minus" on the square root. (Wink)