Find Domain Of Inverse Function

Explanation:

First, we must determine whether $f^{-1} (x)$ exists.

A quadratic function has a parabola as its graph; this graph decreases, then increases (or vice versa), with a vertex at which the change takes place.

$f^{-1} (x)$ exists if and merely if, if f(a) = f(b) , and then a = b - or, equivalently, if there doesnotexist and such that $a \ne b$ , merely. This will happen on any interval on which the graph of constantly increases or constantly decreases, but if the graph changes direction on an interval, there will be $a \ne b$ such that on this interval. The central is therefore to make up one's mind whether the interval to which the domain is restricted contains the vertex.

The-coordinate of the vertex of the parabola of the office

$f(x) = ax^{2}+ bx+c$

is $x = - \frac{b}{2a}$ .

The-coordinate of the vertex of the parabola of $f(x) = x^{2}- 4x - 4$ can exist found by setting a = 1, b = -4 :

$x = - \frac{-4}{2 \cdot 1} = \frac{4}{2} =2$ .

The vertex of the graph ofwithout its domain restriction is at the point with -coordinate 2. However, $2 \notin \left ( -\infty , 0 \right ]$ . Therefore, the domain at which f(x) is restricted does not include the vertex, and $f^{-1} (x)$ exists on this domain.

To determine the changed of f(x) , first, rewrite f(x) in vertex form

f(x) = a (x-h)+k

a = 1 , the aforementioned as in the standard form.

The graph of, if unrestricted, would have a vertex with-coordinate two, and-coordinate

$f(2) = 2^{2}- 4 \cdot 2 - 4 = 4 - 8 - 4 =-8$ .

Therefore, h = 2, k=8 .

The vertex class of f(x) is therefore

$f(x)= (x - 2)^{2}- 8$

To find $f^{-1}(x)$ , first replace f(x) with:

$y= (x - 2)^{2}- 8$

Switch and:

$x= (y - 2)^{2}- 8$

Solve for. Kickoff, add eight to both sides:

$x+ 8 = (y - 2)^{2}- 8 + 8$

$x+ 8 = (y - 2)^{2}$

Take the square root of both sides:

$y - 2 = \pm \sqrt{ x+ 8 }$

Add together 2 to both sides

$y - 2+ 2 = \pm \sqrt{ x+ 8 } + 2$

$y = 2 \pm \sqrt{ x+ 8 }$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)= 2 \pm \sqrt{ x+ 8 }$

Either $f^{-1}(x)= 2 + \sqrt{ x+ 8 }$ or $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$

The domain of f(x) is the set up of nonpositive numbers; this is consequently the range of $f^{-1}(x)$ . $2 + \sqrt{ x+ 8 }$ can only have positive values, so the only possible pick for $f^{-1}(x)$ is $f^{-1}(x)= 2 - \sqrt{ x+ 8 }$ .