# Math Help - Inverse functions problem

1. ## Inverse 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 $3x^2 - 4 - y = 0$
Solving this equation:

$=> \frac{4 \pm \sqrt{16 + 12y}}{6}$

$= \frac{4 \pm \sqrt{2} \sqrt{4 + 3y}}{6}$

$= \sqrt{4 + 3y}$ or $\frac {\sqrt{4+3y}}{3}$

Either of them is wrong.. The answer should be $\sqrt{\frac{1}{3}(x + 4)}$

2. Originally Posted by struck
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 $3x^2 - 4 - y = 0$
Solving this equation:

$=> \frac{4 \pm \sqrt{16 + 12y}}{6}$ Mr F says: This is the solution (for x) to ${\color{red}3x^2 - 4{\color{blue}x} - y = 0}$, but that's not the equation you've got ....

$= \frac{4 \pm \sqrt{2} \sqrt{4 + 3y}}{6}$

$= \sqrt{4 + 3y}$ or $\frac {\sqrt{4+3y}}{3}$

Either of them is wrong.. The answer should be $\sqrt{\frac{1}{3}(x + 4)}$
The inverse is found by solving $x = 3y^2 - 4$ for $y$, where $y = f^{-1} (x)$. Start by adding 4 to both sides.

3. Originally Posted by struck
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 ...
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:

. . . . . $f(x)\, =\, 3x^2\, -\, 4$

. . . . . $y\, =\, 3x^2\, -\, 4$

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

. . . . . $0\, =\, 3x^2\, +\, (-4\, -\, y)$

. . . . . $x\, =\, \frac{0\, \pm\, \sqrt{(0)^2\, -\, 4(3)(-4\, -\, y)}}{2(3)}$

. . . . . . . . $=\, \frac{\pm \sqrt{12(4\, +\, y)}}{6}$

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.