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Switch x and f(x) and solve for f(x).
First simplify e(f(x)).
The graphs are reflection in what line?
To draw the inverse, switch the x- and y-coordinate of some points on f(x).
e(x)=(x-1)^2-5
Result: The functions undo each other.
Function: e(f(-4))=- 4
Explanation: See solution.
See solution.
Diagram:
Domain f(x): x ≥ -5
Range f(x): y ≥ 1
Domain: e(x): x ≥ 1
Range: e(x): y ≥ -5
Let's first find the inverse function. The first step in doing so is switching x and f(x) in f(x).
Rearrange equation
LHS-1=RHS-1
LHS^2=RHS^2
LHS-5=RHS-5
Replace f(x) with e(x)
Let's first create the composite function e(f(x)) by substituting the function f(x) into e(x).
e( f(x))=(( 1+sqrt(x+5))-1)^2-5
Having simplified e(f(x)), we see that it equals x. Therefore, the value of e(f(-4)) must be -4. e(f( -4))= -4
If we draw both graphs on the same set of axes, they will be reflections of each other in y=x. This is true for all functions and their inverse.
The graph goes from -5 and to the right on the x-axis. It also goes from 1 and up on the y-axis. With this information, we can identify the range and domain of f(x).
Just like the inverse swaps the x- and y-coordinates of all points on the function, it also switches the domain and range. We can see this in the diagram as well. Domain e(x):& x≥ 1 Range e(x):& y ≥ -5