Inverse+Functions

Inverse Functions

 * What restriction on the domain of sec(x) is made, so that the function f(x)=sec(x) has an inverse? What are the domain and range of f(x)=arcsec(x)? Provide a sketch of the graph of f(x).**

In order for the f(x)=sec(x) to have an inverse, the domain of the graph must be limited to [0, pi (3.14...)]

Domain: ( negative infinity, -1] U [1, infinity) Range: [0, pi/2) U (pi/2, pi]


 * What restriction on the domain of sin(x) is made, so that the function f(x)=sin(x) has an inverse? What are the domain and range of f(x)=sin^-1(x)? Provide a sketch of the graph of f(x). (by Allison Marron)**

sin(x) does not normally have an inverse because it is not one to one, but a restriction can be made on the domain of sin(x) so that the function f(x)=sin(x) has an inverse.

-want to restrict the domain, but need to keep the range at [-1,1] -restrict domain of sinx to [-pi/2,pi/2]



domain: [-pi/2, pi/2] range: [-1,1]

This makes it one to one so we are now able to take the inverse of it.

y=sin^-1x = y=arcsinx = x=siny

The graph of arcsinx is a reflection of the graph of sin (restricted to [-pi/2, pi/2]) about the lin y=x

Graph of arcsinx:

**How do you find the inverse of a function?** In the equation switch the "x" and "y" and then solve for y. For example: y = 3x-2 1﻿) x = 3y-2 2) x + 2 = 3y 3) (x+2)/3 = y Another way (if only presented with a graph and some points) is to make each point (x,y)--> (y,x) For example: (3,0) --> (0,3) (5, 1.5) --> (1.5, 5) **What restriction on the domain of cos(x) is made, so that the function f(x)=cos(x) has an inverse? What are the domain and range of f(x)=cos^-1(x)? Provide a sketch of the graph of f(x).(Evan HK)** The cosine graph must be restricted to [0, π] for it to be one to one and have an inverse.    Domain: [-1,1] Range[0, π]

Prove arcsec(x) = arcos(1/x) arcsec(x) = arcos(1/x) sec[arcsec(x) = arcos(1/x)] x = sec[arcos(1/x)] x = 1/{cos[arcos(1/x)]} x = 1/(1/x) x = x   Therefore arcsec(x) = arcos(1/x)