Inverse+Functions

Inverse Functions
Amelia Grant-Alfieri: inverse of cot(x) First, we must restrict the domain of cot(x) to one period (0, pi) because otherwise it will not pass the horizontal line test. The domain becomes the inverse's range: (0, pi) The range becomes the inverse's domain: (-infinity, infinity) Switch the x and y from the original equation, in addition to the x and y values of the points to find the new points. Here is the cot(x) - domain restriction in red: Here is the inverse of cot(x):

Isa Cuervo: how to graph the inverse of a function using just the graph of the original function... if not provided with the equation you can't just switch the X and Y as you would usually do. In this case, you would just switch the x and y values of each point on the graph [ (x,y) --> (y,x) ] Example: y= points include: (0,0) ; (1,1) ; (2,root2) ; (4, 2). new points on the inverse graph would be : (0,0) ; (1,1) ; (root2, 2) ; (2,4)

__Inverse of Tangent Function: Jeremy Lash__

Here is the Tangent function where x can't equal (pi/2)n (where n is any integer). First, we must flip flop the x and y values of the tan(x) function within the domain of (-pi/2. pi/2) to receive our inverse tangent function graph. We do this because the inverse of a function must be 1 to 1, therefore causing us to restrict the range of the inverse function from an infinite amount of numbers, to (-pi/2, pi/2). You can also think of it as taking the domain of tan(x) and restricting it in order to receive a range that correlates to a 1 to 1 function.

Therefore, the inverse of a tangent function looks like this: Where x = (-infinity, infinity) and y =(-pi/2, pi/2)

**Inverse csc graph, by Max Weiss**: Q: What restriction on the Domain of csc(x) is made, so that the function f(x)=csc(x) has an inverse? What is the Domain restriction of f(x)=arccsc(x)? A: Basically the graph of y=csc(x) looks like the graph of y=sin(x) with parabolas at each high point and low point facing towards infinity and negative infinity respectively. So, the points ((pi)/2,1) and ((-pi)/2,1) would exist as vertices, and the asymptote x=0 in the middle of them. When the inverse is taken, the parabolas are put on their sides going along the y- axis with y=0 being an asymptote, and the points (1,(pi)/2) and (-1,(-pi)/2) being the new vertices. However, this can not be a function due to it not being able to pass the vertical line test (more than one y coordinate for an x coordinate). Therefore we must restrict the Domain of f(x)=csc(x) to have X less than of equal to negative one greater than or equal to one. This fixes the problem with the inverse not being one-to-one.

Normal->Inverse  Explain why sin (sin-1(∏)) ≠ ∏ Alec Schwartzman Due to rules of the domain of inverse sine which state that the x value must lie between -1 and 1, it is impossible to take the sin inverse of pi, therefore making the statement above true. Since all the trigonometric functions do not pass the horizontal line test, it is necessary to restrict the domain of the original trigonometric function so an inverse function will exist. So when the sine function is restricted to pass the horizontal line test its domain changes to [-∏/2, ∏/2] and its range stays at [-1, 1]. Therefore in order for sine inverse to exist, the domain of sine needs to be restricted to [-1,1]. When the x and y switch roles, this causes the domain and range switch too. So the domain of sine inverse is [-1,1], disallowing ∏ as a possibility in the equation.

Explain why Cos-1(Cos(4∏/3)) ≠ 4∏/3 and why Cos(Cos-1(4∏/3) ≠ 4∏/3: Katrina Tomas In order for F-1(Fx) = X, x must be in the domain of f(x) as well as the range of f-1(x). The domain of cosx is all real numbers but the range of cos-1(x) is [0,∏], therefore cos-1(cosx) will equal x when x falls into the domain of [0,∏]. 4∏/3 is larger than ∏, thus the solution for this cannot be 4∏/3. In order to find the solution, find the cos(4∏/3) which is -1/2, then plug this back into cos-1(-1/2) which is 2∏/3. Similarly, for cos(cos-1(x)) to equal x, x must be in the domain of cos-1(x), [-1,1] and in the range of cosx [-1,1]. Because 4∏/3 is not in the domain of cos-1(x), cos(cos-1(4∏/3)) has no solution. In short, if you encounter a problem like F-1(f(x)), first check to see if x falls into the domain of the inside, in this case the inside is F(x). If it does, then there is a possible solution, if it does not, then there is no solution. If there is a possible solution, you can just solve individually from there.